how to paste part of the file name to the content of file?

plain bash

for file in *.bam_dp; do 
    contents=$(< "$file")
    echo "$contents ${file%%.*}" > "$file"
done

---

for multi-line files, still can be accomplished with plain bash

for file in *.bam_dp; do 
    mapfile -t contents < "$file"
    printf "%s\n" "${contents[@]/%/ ${file%%.*}}" > "$file"
done

notes:

• the mapfile command reads the file into an array of lines.
• the ${var/pattern/string} parameter expansion does a search-and-replace on the variable value. (documented in the manual)
  • if pattern starts with % the pattern is anchored at the end of the string. Here, I'm matching the empty pattern at the end of the string.
  • the variable can be an array expansions, in which case the replacement occurs for each array element.

Frankly, this approach is too clever, and I'd go for something more obvious.

Use a loop:

#!/bin/bash

shopt -s nullglob
for file in ???????.mapped.*bam_dp; do
  [[ -f "$file" ]] || continue
  id=${file%%.*}              # grab the ID from file name
  sed -i "s/$/ $id/" "$file"  # modify the file in-place
done

Remove .* from $ARGV then append \t $ARGV to the file:

perl -i -pe '$ARGV=~s/\..*//; s/$/\t$ARGV/;' NA*

Glenn's solution is most likely faster to run:

perl -i -lpe '$_ .= " " . substr($ARGV,0,index($ARGV,"."))' NA*

though if each file is only a single line, most of the time will be seeking on the drive.