It's all perfectly logical really. The first thing to notice is that the initialisation of $c makes no difference to the outcome:

perl -e "$c=qq(z);$d=qq(d); $c = $d++; $d = $c++; print $c,$d;"
ed

perl -e "$d=qq(d); $c = $d++; $d = $c++; print $c,$d;"
ed
[download]

So,

1. $d is set to 'd'
2. $c is set to $d ('d')
3. $d is incremented ('e')
4. $d is set to $c ('d')
5. $c is incremented ('e')
6. 'ed' is printed.

In the end, the whole thing reduces to:

perl -e "$d=qq(d); $c = $d; $d = $c++; print $c,$d;"
ed
[download]

or even just

perl -e "$c = $d = qq(d); $c++; print $c,$d;"
ed
[download]

Which looks a lot less mysterious :)