I'm having trouble understanding your code.
The mkdir utility doesn't write anything to standard output and the find utility and the cp utilities it invokes don't read anything from standard input. Why do you have a pipeline connecting those two commands?

And, in the 1st post in this thread you had the single pipeline:

Code:

find /home/directory/origin ./ -iname "*.mp3" -exec sha256sum {} \; | 
sort -k 64 > /tmp/runner.txt | cp -nr /home/directory/origin /home/directory/destination | cat -n /tmp/runner.txt | cut -d " " -f1; exit 0;

where sort (with the arguments presented) does not write anything to standard output, cp (with the arguments presented) does not read anything from standard input, cp does not write anything to standard output, and cat (with the arguments presented) does not read anything from standard input. So, in this pipeline, the find, sort, cat, and cut have absolutely no affect on how the cp command in that pipeline behaves.

I don't have a sha256sum utility on my system. So I can't see what the 64th field is in the output it produces, and I don't understand why sorting on the 64th field is important when it looks like all that happens is that you print the 1st field to the terminal after sorting the data into a file, reading the file back in, and piping it into cut. There is nothing here that makes any attempt to compare the checksum for any file to the checksum of any other file.

If you are trying to copy files from multiple source directories into a single target directory, why does a checksum on a source file or on a target file matter? Only one file with a given name can exist in your target directory. Why does a checksum matter in deciding whether or not you want to replace an existing file in your destination directory?