I've been thinking on the same lines too. :) @vizzard110
I think for a given A there is only one possible P(A).
2 different A's cannot map to a single P(A) so it is one-to one.
And for a given set of A's we have a finite set of P(A)'s, so if we "assume" co-domain (they haven't defined what needs to be taken as co-domainπ₯) to be the set of all P(A)'s then we have co-domain = range, and hence it is onto.
So I got the answer as option 3) f is both one-to-one and onto.