1. When using capture groups, refer to them via $1, $2 etc.
2. To evaluate code in the replacement part of substitution, use /e .
3. You probably need + to match numbers > 9:

   s/(\w+)/num2en($1)/ge
   [download]

   ($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
   }map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
   [download]

if(isdigit (\w)
[download]

But you probably don't need this conditional anyway, since your substitution on the next line can be modified to match only numbers, with something like this (untested):

s/(\d+)/num2en($1)/ge;
[download]

Indeed! And at first sight I was surprised not to see the usual unmatched bracket message, though most likely it is giving up further checking before the match brackets check when finding \w to be a bareword.

One world, one people

choroba gave you the solution: but why use isdigit when you can already match directly digits with \d classe in regexes?

I gave a try to this never used by me module ending with the below oneliner: For sure is not the best thing ever seen ( PS indeed! see tremendously wise AnomalousMonk below) but you can play with it (be aware of the win32 double quotes!):

echo 1st rule: 2nd, 3rd and 4th floor must be free in 5 min. Not in 6.
+ Call 911 for emergencies   |

perl -MLingua::EN::Numbers="num2en,num2en_ordinal" -pne "s/(\d+)([st|n
+d|rd|th])+/num2en_ordinal($1)/ge;s/(\d+)/num2en($1)/ge"

first rule: second, third and fourth floor must be free in five min. N
+ot in six. Call nine hundred and eleven for emergencies
[download]

You can use MO=Deparse to see the above a bit expanded:

perl -MO=Deparse  -pne "s/(\d+)([st|nd|rd|th])+/num2en_ordinal($1)/ge;
+s/(\d+)/num2en($1)/ge"

LINE: while (defined($_ = <ARGV>)) {
    s/(\d+)([st|nd|rd|th])+/num2en_ordinal($1);/eg;
    s/(\d+)/num2en($1);/eg;
}
continue {
    die "-p destination: $!\n" unless print $_;
}
-e syntax OK
[download]

PPS maybe this is better regex (i'm so rusty..), the char class [..] was totally misplaced, but being all look around assertions zero width ones ( $& not filled ) a simple capturing group seems to work:

 
s/(\d+)(st|nd|rd|th)/num2en_ordinal($1)/ge
[download]

L*

s/(\d+)([st|nd|rd|th])+/num2en_ordinal($1)/ge;

The  | ordered alternation metacharacter is not meta in a character class. The character class [st|nd|rd|th] is equivalent to [stndrh|] (update: because repeated characters have no special significance in a class). The quantified capture group ([st|nd|rd|th])+ matches any of the characters in the [stndrh|] class one or more times and captures the last such character matched. Try matching against  '2ds' '33hn' '123|s' and a few other such strings. You may want something like the true alternation (?:st|nd|rd|th) possibly supported by some look-around assertions.

c:\@Work\Perl\monks>perl -wMstrict -le
"my $s = '1ds rule: 22hn, 3rd and 123|sdnfloor';
 ;;
 printf qq{'$&' } while $s =~ m{(\d+)([st|nd|rd|th])+}xmsg;
"
'1ds' '22hn' '3rd' '123|sdn'
[download]

Update: Maybe something like this:

c:\@Work\Perl\monks>perl -wMstrict -le
"my $rx_ordinal_indicator =
   qr{ (?<! [[:alpha:]]) (?: st | nd | rd | th) (?! [[:alpha:]]) }xms;
 ;;
 my $s = '1ds rule: 22hn, 2 then 3rd and 44 th, 2  nd of the 123|sdnfl
+oor';
 ;;
 printf qq{'$&'  } while $s =~ m{ (\d+) \s* $rx_ordinal_indicator }xms
+g;
"
'3rd'  '44 th'  '2  nd'
[download]

(Ordinarily, I wouldn't use  $& and friends, but it's convenient for this example.)

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