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Let's suppose that $p >3$ (otherwise, the groups is solvable in any case). I also work with $G = {\rm SL}(2,\mathbb{Z}/p^n \mathbb{Z})$, but there is an obvious correspondence between what happens for ${\rm PSL}$ and what happens for ${\rm SL}$. The group $G$ has a largest normal $p$-subgroup (denoted, as is customary for group theorists, by $O_p(G)$), and a central subgroup $Z$ of order $2$. $G/Z \times O_{p}(G)$ is isomorphic to the simple group ${\rm PSL}(2,\mathbb{Z}/p\mathbb{Z})$. Furthermore, $H = O_{p}(G)$ is the set of elements of $G$ which are congruent (entrywise) to the identity $(mod p)$. More generally, we have a series of obvious normal subgroups $$ 1 = H_n < H_{n-1} < \ldots < H_{1} = H < G,$$ where $H_{i}$ is the set of elements of $G$ congruent (entrywise) to the identity $(mod p^i)$. For each $j$, $H_{j}/H_{j+1}$ is an elementary Abelian $p$-group (that is, an Abelian $p$-group of exponent $p$). To see this, note that for $a,b \in H_j$, we have $ab -ba = (a-I)(b-I) - (b-I)(a-I) \equiv 0$ (mod $p^{2j})$, so that $I - a^{-1}b^{-1}ab \equiv 0$ (mod $p^{2j}$). Hence $[H_{j},H_{j}] \subseteq H_{2j} \subseteq H_{j+1}$ and $H_{J}/H_{j+1}$ is Abelian (in fact $H_{j}/H_{2j}$ is Abelian). Also , for $a \in H_j$, we have $a^{p} - I = (a-I)(a^{p-1}+ \ldots a^{p-2} + \ldots + I) \equiv 0$ (mod $p^{j+1})$, so that $a^p \in H_{j+1}$. Each of the factor groups $H_{j}/H_{j+1}$ has the structure of a module for the group ${\rm SL}(2,\mathbb{Z}/p\mathbb{Z})$ over the field of $p$-elements. The precise structure of these modules is of great interest (see, for example, the work of N. Boston).