Let $G$ be a profinite group which fits into the following short exact sequence: $$ 1\rightarrow N\rightarrow G \rightarrow K\rightarrow 1 $$ Assume that $N$ is a pro-$p$ group and that $K$ is topologically finitely generated. Note that $K$ acts naturally by conjugation on $N^{ab}$ and thus we way view $N^{ab}$ as a $\mathbf{Z}_p[[K]]$-module.
Q: If $N^{ab}$ is a finite type $\mathbf{Z}_p[[K]]$-module, does it follow that $G$ is topologically finitely generated?
So here is my comment again with slightly more details. Let $Y$ be a finite subset of $G$ such that its image generates $K$. As I was told by Yves, finite type means finitely generated. Thus, let $X$ be a finite subset of $N$ such that $N^{ab}$ is generated by its image as a $\mathbb{Z}_p[[K]]$-module. Then we claim that $H$ the subgroup generated (topologically) by $X \cup Y$ is equal $G$. Since $Y \subseteq H$ we have that $HN/N \cong K$.
Therefore, $HN=G$. So it suffices to show that $N \leq H$. By definition, the action of $K$ on $N^{ab}$ is via the action of $G$. Since $N$ acts trivially on $N^{ab}$ we have that $H$ action on $N^{ab}$ is the same as $HN=G$ action. Hence, the image of $Y^{H}$ generates $N^{ab}$ as a $\mathbb{Z_p}$-module. So $Y^{H}$ generates $N/\Phi(N)$ as a group, where $\Phi(N)=[N,N]N^{p}$ is the Frattini subgroup of $N$. We deduce that $Y^{H}$ generates $N$ (topologically) and $N \leq H$ as we wanted.
Note that it suffices to ask that $N/([N,N]N^{p})$ (EDIT: see comments below) is finitely generated as a $K$-module.
