Note that this equality is as topological groups, where Hom is endowed with uniform convergence on compact subsets.

Every such homomorphism is zero on $p^n$ for some $n$. Hence the given homomorphism group $G$ can be written as $\bigcup G_n$, where $G_n$ is the set of homomorphisms that vanish on $p^n$. Note that $G_n$ is open.

By definition, $$G_0=\mathrm{Hom}(\mathbf{Q}/\mathbf{Z},\mathbf{Z}(p^\infty))=\mathrm{Hom}(\bigoplus_\ell\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=\mathrm{End}(\mathbf{Z}(p^\infty)),$$ because for prime $\ell\neq p$ we have $\mathrm{Hom}(\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=0$. It is classical that $\mathrm{End}(\mathbf{Z}(p^\infty))$ is isomorphic to $\mathbf{Z}_p$ as a topological ring (details upon request).

Hence each $G_n$ is isomorphic to $\mathbf{Z}_p$. Multiplication by $p$ maps $G_n$ into $G_{n-1}$. Also, mapping $f$ to $x\mapsto f(p^{-1}x)$ maps $G_{n-1}$ into $G_n$ and is an inverse to multiplication by $p$. So indeed $pG_n=G_{n-1}$, and we deduce the result.

Note that this equality is as topological groups, where Hom is endowed with uniform convergence on compact subsets.

Every such homomorphism is zero on $p^n$ for some $n$. Hence the given homomorphism group $G=\mathrm{Hom}(\mathbf{Q},\mathbf{Z}(p^{\infty}))$ can be written as $\bigcup G_n$, where $G_n$ is the set of homomorphisms that vanish on $p^n$. Note that $G_n$ is open.

By definition, $$G_0=\mathrm{Hom}(\mathbf{Q}/\mathbf{Z},\mathbf{Z}(p^\infty))=\mathrm{Hom}(\bigoplus_\ell\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=\mathrm{End}(\mathbf{Z}(p^\infty)),$$ because for prime $\ell\neq p$ we have $\mathrm{Hom}(\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=0$. It is classical that $\mathrm{End}(\mathbf{Z}(p^\infty))$ is isomorphic to $\mathbf{Z}_p$ as a topological ring (details upon request).

Hence each $G_n$ is isomorphic to $\mathbf{Z}_p$. Multiplication by $p$ maps $G_n$ into $G_{n-1}$. Also, mapping $f$ to $x\mapsto f(p^{-1}x)$ maps $G_{n-1}$ into $G_n$ and is an inverse to multiplication by $p$. So indeed $pG_n=G_{n-1}$, and we deduce the result.

Edit: consider $\mathbf{Q}_p$ acting on itself as its own endomophism group (as topological group). Restricting to $\mathbf{Q}$ at the origin, and modding out by $\mathbf{Z}_p$ at the target yields a topological group homomorphism $\phi:\mathbf{Q}_p\to\mathrm{Hom}(\mathbf{Q}_p/\mathbf{Z}_p)$ (where the latter Hom is in topological groups). This homomorphism $\phi$ is injective: indeed, an element $f$ in the kernel of $\phi$ would, in restriction to $\mathbf{Q}$, satisfy $f(\mathbf{Q}_p)\subset\mathbf{Z}_p$. Since $\mathbf{Q}_p$ is divisible it has no nontrivial finite quotient and hence no nonzero homomorphism into any profinite group. So $f=0$.

Keeping in mind that there is a canonical isomorphism $\mathbf{Z}(p^\infty)\to\mathbf{Q}_p/\mathbf{Z}_p$, since we previously obtained that $\mathrm{Hom}(\mathbf{Q}_p/\mathbf{Z}_p)\simeq\mathbf{Q}_p$ as topological group, we deduce that $\phi$ is a isomorphism.

I should also add that it's important to phrase the result in terms of topological groups. Indeed, as an abstract group, $\mathbf{Q}_p$ is just isomorphic to $\mathbf{Q}^{(c)}$ (and also to $\mathbf{R}$).