my $saved_pass = 'bd0e9f94ce671b3cdd13081fa5a8b32f9ccd9ebf';
my $eword = 'bd0e9f94ce671b3cdd13081fa5a8b32f9ccd9ebf';

my $passwd_check;
$saved_pass eq $eword  ?  $passwd_check = "good"
                       :  $passwd_check = "fail";
                       
print $passwd_check;
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fail

my $saved_pass = 'bd0e9f94ce671b3cdd13081fa5a8b32f9ccd9ebf';
my $eword = 'bd0e9f94ce671b3cdd13081fa5a8b32f9ccd9ebf';

my $passwd_check = $saved_pass eq $eword  ?  "good" : "fail";
                       
print $passwd_check;
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good

I have do admit that I don't really get why. Anyone care to explain?
update: Got it (from perlop):

Because this operator produces an assignable result, using assignments without parentheses will get you in trouble. For example, this:

$a % 2 ? $a += 10 : $a += 2
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Really means this:

(($a % 2) ? ($a += 10) : $a) += 2
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Rather than this:

($a % 2) ? ($a += 10) : ($a += 2)
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so what you wrote really means:

($saved_pass eq $eword  ?  $passwd_check = "good"  :  $passwd_check ) 
+= "fail";
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perlop also says that you can assign to the ternary operator if both arguments between the ? and : are valid lvalues.
So the code comes down to:

if ($saved_pass eq $eword) {
        $passwd_check = "fail"; 
}else{
        $passwd_check = "fail";
}
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In words: assign "fail" to $passwd_check or $passwd_check :-)