perl -MO=Deparse

First, the obfuscated code:

#!/usr/bin/perl
                                                                      
+          
$ == _ eq _;
$ -= ( ) = qw;      ( );;
$ *= $ -* $ -* $- ;
$ /= $ -+ $ *; _:
$ ".= chr $ /+ $ *** $ -;
$ /= $ / >> $ = || $ ** $ -;
$ /= $ / &~ $ */ $ - and goto _;
                                                                      
+          
print $ "," 
"
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Pretty confusing! But let's run it through -MO=Deparse, and we get:

$= = '_' eq '_';
$- = () = ('(', ')');
$* = $- * $- * $-;
$/ = $- + $*;
_: $" .= chr $/ + $* ** $-;
$/ = $/ >> $= || $* * $-;
$/ = $/ & ~$* / $- and goto _;
print $", "\n";
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Still pretty confusing, but at least now I can parse it a little better.

$= = '_' eq '_';
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$= normally defaults to 60 as the current page length of the selected output channel, but this just sets it to 1.

$- = () = ('(',')');
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$- = () = qw; ( );; is like $- = () = ('(',')');

('(',')') is a two element list, and the =()= idiom Perl Idioms Explained - my $count = () = /.../g shows that $- receives the count of elements in the list, in other words 2.

$* = $- * $- * $-;
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Equivalent to $* = 2 * 2 * 2, or 8.

$/ = $- + $*;
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Sets $/ to 10.

_: $" .= chr $/ + $* ** $-;
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_: is a line label, and I thought (as Errto noted above) it had to be at the begining of a line. $=, the list separator for list values interpolated into a double quoted string, defaults to a single space. The statement is equivalent to $" =  " " . chr 10 + 8 ** 2 , and ** binds tighter than + , therefore $" = " " . chr 10 + 64, or rather
$" = " J".

$/ = $/ >> $= || $* * $-;
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This is: $/ = 10 >> 1 || $* * $-;

10 >> 1 will evaluate true (non-zero), so the right half doesn't gets evaluated, and $/ is subsequently set to 5.

$/ = $/ &  ~$*  / $-  and goto _;
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This can be written as $/ = 5 & ~ 8 / 2 , or a bit more clearly, $/ = 5 & ( ( ~ 8 ) / 2 ). This has the net effect of setting $/ to 1.

Going to the "_" line label, $= (currently " J") has chr 1 + 64, or "A" appended to it. then 1 >> 1 evaluates to 0, so this time the right hand of the expression is evaluated, and $/ is set to 8 * 2, or 16. Then $/ = $/ & ~ $* / $- can be restated as $/ = 16 & ( ( ~ 8 ) / 2 ), and sets $/ to 16.

Back to the "_" line label. $= (currently " JA") has chr 16 + 64, or "P" appended to it. 16 >> 1 is 8, so $/ is set to 8. and 8 & ( ( ~ 8 ) / 2 ) evaulates to 8.

Back to the "_" line label, $= (currently " JAP") has chr 8 + 64, or "H" appended to it. Then 8 >> 1 is 4, so $/ is set to 4. And what I think is the really clever "bit" ;) is that now 4 & ( ( ~ 8 ) / 2 ) evaluates to 0, so and goto  _; is skipped, proceeding to:

print $", "\n";
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... which I think is pretty self-explanatory.