How to extract the part of filename with specific pattern?

Perhaps:

for file_name in *0728*dat
do
  printf '%s\n' "${file_name%[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]_[0-9][0-9]_[0-9][0-9]_[0-9][0-9].dat}"
done

This strips from the end of each filename the pattern of: 8 digits, underscore, 2 digits, underscore, 2 digits, underscore, 2 digits, then .dat.

using sed:

ls -a *0728*dat | sed 's/[0-9].*//'
amnbmnb_kjhkj_
amnbmnbm_jnjmnm_sd_
njnkjnjk_AbnBCBB_DE_

or

ls -a *0728*dat | sed "s/[0-9]\{8\}_[0-9]\{2\}_[0-9]\{2\}_[0-9]\{2\}\.dat$//"

The following requires bash 4.4 or newer, GNU find and GNU sed (or, at least a find and a sed that support NUL as the output/input separator):

$ mapfile -d '' files < <(find . -maxdepth 1 -type f -iname '*0728*.dat' -print0 | 
    sed -z 's:^\./::; s/[0-9][0-9_]\+.dat//')

This populates a bash array (files) with the matching filenames in the current directory after they've had the leading ./ and the date & time & .dat extension stripped by sed. It uses NUL as the record (i.e. filename) separator to avoid any potential issues with spaces, line-feeds, shell meta-characters, etc in any of the filenames.

find is used because you should never use the output of ls as input to other programs or as arguments on another program's command-line - it is unsafe and unreliable. See Why not parse ls (and what to do instead)?

The find command can, of course, be modified to find files matching different patterns, or in sub-directories, etc.

Example:

$ touch amnbmnbm_jnjmnm_sd_07282019_14_13_17.dat amnbmnb_kjhkj_07282019_11_23_22.dat \
    njnkjnjk_AbnBCBB_DE_07282019_07_09_04.dat

$ mapfile -d '' files < <(find . -maxdepth 1 -type f -iname '*0728*.dat' -print0 | 
    sed -z 's:^\./::; s/[0-9][0-9_]\+.dat//')

$ typeset -p files
declare -a files=([0]="amnbmnbm_jnjmnm_sd_" [1]="amnbmnb_kjhkj_" [2]="njnkjnjk_AbnBCBB_DE_")

$ printf '%s\n' "${files[@]}"
amnbmnbm_jnjmnm_sd_
amnbmnb_kjhkj_
njnkjnjk_AbnBCBB_DE_