= Solution
Since $N$ is normal in $G$, it is normal in $H=\langle x,N\rangle$, and $H/N$ is a <cyclic group>. More specifically, quotienting $H$ by $[N,H]$ makes the image of $N$ central. The quotient is generated by this central image and the single element $x$, and is therefore abelian. Hence
$$
[H,H]\leq[N,H]\leq[N,G]\leq N^p\leq H^p.
$$
\b[Thus $H$ is a powerful p-group.] Merely saying that $H/N$ is cyclic would give only $[H,H]\leq N$; centralizing $N$ is the extra step needed here.
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