Try set correct data type for returned data.

$.ajax({
  url: 'script_admin-add-category.php',
  method: 'POST',
  data: {lExpensesId: lcl_ExpensesId},
  dataType: 'text',
  success: function (data) {
    if (data === 'YES') {
      alert('EMAIL')
    } else {
      alert(data)
    }
  }
})

@J Salaria as i understood your question you are having problem with jquery AJAX and PHP code as you are not getting you desired result. There are different ways to send the data through jquery ajax which i will be explain in detail.

$_POST["lExpensesId"] are you getting this ID from a HTML <form> ?.Because here i'll be showing you 3 different practiced ways to send data through ajax..

NOTE: YOUR CODE IS VULNERABLE TO SQL INJECION. I'LL BE ALSO SHOWING YOU THE METHODS TO OVERCOME.IF YOU WANT TO LEARN MORE ABOUT SQL INJECTION CLICK ON THIS LINK SQL INJECTION LINK

HTML FORM CODE :

<form action="" id="send_lExpensesId_form" method="post">
    <input type="text" name="lExpensesId" id="lExpensesId" >
    <input type="submit" name="submit" >
</form>

FIRST WAY FOR SENDING DATA THIS IS THOUGH HTML <FORM>

 <script>
    $(document).ready(function(){
        $("#send_lExpensesId_form").submit(function(e){
            e.preventDefault();

            var form_serialize = $(this).serialize();

            $.ajax({
                type:'POST',
                url:'script_admin-add-category.php',
                data:form_serialize,
                success:function(data){
                    if(data == "YES"){
                        alert("EMAIL");
                    }else{
                        alert(data);
                    }
                }
            });
        });
    });
</script>

SECOND WAY FOR SENDING DATA THIS IS THOUGH HTML <FORM>

    <script>
    $(document).ready(function(){
        $("#send_lExpensesId_form").submit(function(e){
            e.preventDefault();

            var form_serialize = new FormData($(this)[0]);


            $.ajax({
                type:'POST',
                url:'script_admin-add-category.php',
                data:form_serialize,
                contentType: false,
                processData: false,
                success:function(data){
                    if(data == "YES"){
                        alert("EMAIL");
                    }else{
                        alert(data);
                    }
                }
            }); 
        });
    });
</script>

THIRD WAY FOR SENDING DATA THIS IS USED WHEN A LINK CLICKED OR TO DELETED THROUGH ID OR CLASS

 <script>
    $(document).ready(function(){
        $("#send_lExpensesId_form").submit(function(e){
            e.preventDefault();

            var lcl_ExpensesId = $("#lExpensesId").val();

            $.ajax({
                type:'POST',
                url:'script_admin-add-category.php',
                data:{lExpensesId:lcl_ExpensesId},
                success:function(data){
                    if(data == "YES"){
                        alert("EMAIL");
                    }else{
                        alert(data);
                    }
                }
            });
        });
    });
</script>

HERE IT THE PHP CODE WITH mysqli_real_escape_string(); AGAINST SQL INJECTION

    <?php
    $servername = "localhost";
    $username = "root";
    $password = "admin";
    $dbname = "demo";

    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $dbname);
    // Check connection
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }

    if(isset($_POST["lExpensesId"])){
        $lExpensesId = mysqli_real_escape_string($conn, $_POST["lExpensesId"]);

        $Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId = '$lExpensesId'";
        if($query_result = mysqli_query($conn, $Lquery)){

            if(mysqli_num_rows($query_result )){
                echo 'YES';
            }else{
                echo "Proceed";
            }
        }else{
            echo "Error".mysqli_connect_error();
        }
    }
?>

HERE IT THE OTHER PHP CODE WITH MYSQLI->PREPARED WHICH IS BETTER AGAINST SQL INJECTION

<?php
    // WITH MYSQLI PREPARED STATEMENT AGAINST SQL INJECTION
    $sql = $conn->stmt_init();
        $Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId =?";
        if($sql->prepare($Lquery)){
            $sql->bind_param('i',$lExpensesId);
            $sql->execute();
            $sql->store_result();

            if($sql->num_rows > 0){
                echo 'YES';
            }else{
                echo "Proceed";
            }
        }
        else
        {
            echo "Error".mysqli_connect_error();
        }
?>

I HOPE YOU GOT ANSWERE FOR YOU QUESTION IF YOU HAVE OTHER DOUBTS FEEL FREE AND COMMENT BELOW

All methods are known and many thanks for assistance. My question is that Why I am not able to get proper return from PHP. Below is my code:

var lcl_ExpensesId = $("#IExpensesId").val();
            //alert(lcl_ExpensesId);
        $.ajax({

                    url     :"script_admin-add-category.php",
                    method  :"POST",
                    data    :{lExpensesId:lcl_ExpensesId},
                    success:function(data){
                        if(data=="ok"){
                            alert("Inserted");
                        }else{
                            alert(data);
                        }
                    }
        });
ob_start();

/------------------FUNCTION TO READ ACCOUNTS DROPDOWN EXPENSES LIST -----------------------/

require_once 'db_config.php';
$newlist    = fxn_CONFIGURATION();
$HOST       = $newlist[0];
$DBNAME     = $newlist[1];
$UNAME      = $newlist[2];
$PSWD       = $newlist[3];
$conn   =   mysqli_connect($HOST, $UNAME, $PSWD, $DBNAME);


    //Check connection
    if(!$conn){
        die("Connection Failed: " .mysqli_connect_error());
    }else{
            if(isset($_POST["lExpensesId"])){

                $lExpensesId    = $_POST["lExpensesId"];

                    $Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId = '$lExpensesId'";
                    $query_result = mysqli_query($conn, $Lquery);

                            if(mysqli_num_rows($query_result) > 0){
                                echo "ok";
                            }else{
                                echo "Proceed";
                            }

            }
    }       
mysqli_close($conn);


ob_flush();

As, i am using this AJAX in my one of input keyup method so whatever I will type, each and everytime, it will execute PHP script. I am having a item as FOOD in databse. When i type "F", I got Proceed, "O" - Proceed, "O" - Proceed, "D" - ok.... When I type D, i should get "Inserted" instead of Ok.... This is my doubt that why i m getting this????

The above problem is resolved by using exit() statement in PHP as I am getting five ↵↵↵↵↵ after my values and it means I am having 5 lines of html without closing ?>. So the best way to resolve the issue is to use exit() in PHP as per need