Q.If $A = \phi$, then $n[P(A)] =$

$P(\phi)=\{\phi\}$, which has exactly $1$ element.

For any set $A$ with $n$ elements, the power set $P(A)$ (the set of all subsets of $A$, including $A$ itself and the empty set) has $2^n$ elements.

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$n[P(A)] = 2^{n(A)}$; with $A=\phi$, $n(A)=0$, so $n[P(A)]=2^0=1$.

The power set $P(A)$ of a set $A$ is the set of all subsets of $A$, and for any set with $n(A)=n$ elements, $n[P(A)]=2^n$.

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Taking $A=\varnothing$ (the empty set) — the standard form of this question — the power set $P(A)=\{\varnothing\}$ has exactly $1$ element.

The power set $P(A)$ of a set $A$ is the set of all subsets of $A$; if $A$ has $n$ elements, $P(A)$ has $2^n$ elements.

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