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$as we wanted. Note that it sufices to ask that$N/([N,K]N^{p})$N/([N,N]N^{p})$ (EDIT: see comments below) is finitely generatd as a $K$-module.
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$as we wanted. Note that it sufices to ask that$N/([N,K]N^{p})$is finitely generatd as a$K\$-module.