Subprocess.Popen() not working when reading file path from csv file

[snippsat]

gives the above-mentioned error code (if the file path has a black space it).

In the new code

print(prg)

Do print(repr(prg)) then you see if there is a space.
It's normal when read from file eg csv to use strip() so there no space at beginning or end.

>>> path = r'C:\code\file.csv '
# Looks ok,but are not
>>> print(path)
C:\code\file.csv 
>>> 
>>> # To see all
>>> print(repr(path))
'C:\\code\\file.csv '
>>> 
>>> # Fix
>>> path.strip()
'C:\\code\\file.csv'