no. the power set of a set, contains ALL subsets of the original set.
for example, if C = {a,b}, then P(C) = {Ø,{a},{b},{a}U{b}} = {Ø,{a},{b},{a,b}} = {Ø, {a},{b},C}
the power set is "bigger" than just {C}.
one way to understand "why" |P(A)| = is to think of all possible functions:
f:A→{0,1}.
for example, one such function is:
f(a) = 1
f(b) = 1
f(c) = 0
f(d) = 1
which corresponds to the set {a,b,d}. see what i did? i just picked a subset of A, in this case, {a,b,d}, and defined f(x) = 1, if x is in {a,b,d}, f(x) = 0, otherwise.
there are such functions (since we have 2 choices for every element f(x), and |A| elements).
if you like, you can think of this as a truth-table for the proposition: "is x in this subset X of A?" a value of 1, means "yes"(true), and a value of 0 means "no"(false).
if you have a set of 3 elements, its power set will have 8 elements. and there are precisely 8 functions from {x,y,z} to {0,1}.