From the documentation on Perl grep():

Evaluates the BLOCK or EXPR for each element of LIST (locally
            setting $_ to each element) and returns the list value con
+sisting
            of those elements for which the expression evaluated to tr
+ue. In
            scalar context, returns the number of times the expression
+ was
            true.
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So you are making a new list, @primes, from the original list, @numbers, including only those elements of @numbers for which the expression returns true.

Your expression,

{
  ( exists $is_prime{$_} and $is_prime{$_} )
  or 
  ( $is_prime{$_} = is_prime($_) )
}
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has two conditions. If either of those returns true for the number from @numbers currently being evaluated, the number is added to @primes.

As grep goes through @numbers, it assigns each one to $_ as it is working with it.

The first condition tests to see whether there is already an entry in the hash %is_prime for the current number:

exists $is_prime{$_}
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If there is not, the second part of the first condition, after the and, is ignored and grep() goes on to the second condition. If the number does exist as a key in the hash,

and $is_prime{$_}
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checks to see whether additionally the value in the hash is true (ie not 0 or undef).

If that first condition returns true, the second condition is ignored. If the first condition is false, the second condition calls the sub is_prime() with the number:

is_prime($_)
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. . . and if the sub returns true, the second condition both adds the number to the hash:

$is_prime{$_} =
# the assigned value is the result of is_prime($_) if that function re
+turned true
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. . . and causes the expression to return true, which adds the number to @primes.

Hope this helps.