 Solution

Since $N$ is normal in $G$, it is normal in $H=⟨x,N⟩$, 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. HenceThus $H$ is a powerful p-group. Merely saying that $H/N$ is cyclic would give only $[H,H]\le N$; centralizing $N$ is the extra step needed here.