This is because of the return value of a subroutine in the absence of a return statement; it is the value of the last statement. In this case, both if and unless have truth values of whatever their argument is if they fail. If they succeed, their truth value is the last statement inside their block. That is:

sub b{unless (!pop) {"unless"} }
sub c{if     (pop)  {"if"    } }
print "failed if (0):         ",c(0), "\n";
print "failed unless (0):     ",b(0), "\n";
print "succeeded if (42):     ",c(42),"\n";
print "succeeded unless (42): ",b(42),"\n";
[download]

prints:

failed if (0):         0
failed unless (0):     1
succeeded if (42):     if
succeeded unless (42): unless
[download]

In this context, it makes sense -- when the conditional is not met, it falls though, and a() returns the last value in the subroutine -- in the "if" case, the value of $c, which is 0. In the case of the unless, it returns the value of !$c, which is 1.

Update: Added two more cases to the example to make it clearer.

perl -pe '"I lo*`+$^X$\"$]!$/"=~m%(.*)%s;$_=$1;y^`+*^e v^#$&V"+@( NO CARRIER'

perl -le 'sub a{my$c=pop;print $c;$c?$c*a($c-1):1}print a(--$ARGV[0])'
+ 5
[download]

   MeowChow                                   
               s aamecha.s a..a\u$&owag.print