On Page 100 of Naive set Theory by Halmos, He asserted that the cardinal number of the power set of a set $A$ is $2^a$ if $card(A)=a$, because "the proof is immediate from the fact that P(A) is equivalent to $2^a$"
However, we defined the cardinal of a set is the least element of the ordinals that are equivalent to $A$. How can we prove that $2^a$ is the least?
There is an abuse of notation here that I think may be confusing you: The notation $2^X$ can refer to the set of all functions $X \to 2$, but when $X$ happens to be a cardinal it can also refer to the cardinality of this set of functions (in which case it is a cardinal by definition.)
Note that even if $X$ happens to be a cardinal the set of functions $X \to 2$ is not literally a cardinal, nor even an ordinal, except in the trivial case $X = 0$.
Letting $A$ be any set and letting $a = \text{card}(A)$, we have $$\text{card}(\mathcal{P}(A)) = \text{card}\big(2^A\big) = \text{card}(2^a)$$ where the first equality comes from considering characteristic functions and the second equality comes from a bijection from $A$ to $a$. If desired, the last step is to drop the "$\text{card}(\cdot )$" from $\text{card}(2^a)$ in accordance with the conventional abuse of notation, giving $\text{card}(\mathcal{P}(A)) = 2^a$.
Remark: For ordinals $\alpha$ and $\beta$ there is another meaning of $\alpha^\beta$, namely ordinal exponentiation. If $\alpha$ and $\beta$ are finite then the two meanings are the same modulo the aforementioned difference between a set of functions and its cardinality.
It can be proved by induction that $\mathcal{P}(\alpha)$ has the same cardinality as $2^\alpha$ in both the ordinal exponentiation and the cardinal exponentiation sense when $\alpha$ is finite, which may be what Newb's comment is alluding to. This equality breaks down at limit stages; ordinal exponentiation is continuous in the exponent and cardinal exponentiation is not.