Contrary to the accepted answer, the language is irregular, but the reasoning is wrong. Instead of requiring m to be a prime, require m to be just any integer ≥ 0. Your argument would stay intact, but the language is actually the regular language $a^n$ for n ≥ 2.
(The accepted answer claims wrongly that the concatenation would always have an even number of a's, when obviously the number of a's can be 2 + p for any odd prime p).
David, I'm deeply disappointed with you. The usual method of demonstrating that an argument is wrong is to show that if the argument was right, it would "prove" something wrong. That's what I did. The argument that the poster gave could be used identically to prove something that is obviously incorrect, therefore the argument is wrong.
The argument given only used that $a^p$ where p is prime was irregular. Nothing about the second language $a^m$, m prime, was used at all. Therefore for the argument to be correct, the concatenation of $a^p$, p prime, with any other language would have to be irregular. Which is incorrect.