Let $\mathbb{Z}_p$ be the ring of $p$-adic integers
I understood that all closed subgroups of $\mathbb{Z}_p$ are $p^n\mathbb{Z}_p$, and multiplicative group of units of $\mathbb{Z}_p$ is $(\mathbb{Z}/(p-1)\mathbb{Z})\times\mathbb{Z}_p$
Then, why the closed subgroup of this unit group is of form $A\times p^n\mathbb{Z}_p$ where $A$ is subgroup of $\mathbb{Z}/(p-1)\mathbb{Z}$?
I don’t have any idea how to find(classify) all closed subgroup of multiplicative group of units of $\mathbb{Z}_p $ Why they must be of form $A\times p^n\mathbb{Z}_p$? Aren’t there any other possibilities?
This does not answer why the subgroups are direct products of subgroups but only which of those (latter) can occur.To see why all subgroups must be direct products of subgroups see the comment by @Torsten Schoeneberg below and in the linked post Tobias Kildetoft´s answer. It follows all subgroups of $\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p$ are of the form $$A\times B$$ where $A$ is a subgroup of $\mathbb{Z}/(p-1)\mathbb{Z}$ and $B$ is a subgroup of $\mathbb{Z}_p$. Since $\mathbb{Z}/(p-1)\mathbb{Z}$ as a topological space is compact, the projection map $$\pi:\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p\rightarrow\mathbb{Z}_p$$ is closed and thus $B$ must be closed in $\mathbb{Z}_p$ whenever $A\times B$ is closed in $\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p$ and thus $B=0$ or $B=p^n\mathbb{Z}_p$ for some $n\in\mathbb{N}_0$.