Multiplication by the prime $p:\Bbb Z\to \Bbb Z$ fits in a short exact sequence $$ 0\to \Bbb Z\xrightarrow{p}\Bbb Z\to \Bbb Z_p\to 0$$ Use this to derive the natural split exact sequence $$0\to H_n(X)/pH_n(X)\to H_n(X;\Bbb Z_p)\to\text{ker}\{p:H_{n-1}(X)\to H_{n-1}(X)\}\to 0$$
This is exercise 4.5.1 of Bredon's topology and geometry textbook. $0\to \Bbb Z\xrightarrow{p}\Bbb Z\to \Bbb Z_p\to 0$ induces $0\to \Delta_{*}(X)\otimes\Bbb Z\to \Delta_{*}(X)\otimes\Bbb Z\to\Delta_*(X)\otimes \Bbb Z_p\to 0$ so a long exact sequence $\cdots\to H_n(X)\to H_n(X)\to H_n(X;\Bbb Z_p)\to H_{n-1}(X)\to H_{n-1}(X)\to\cdots$. Then by first isomorphism theorem, $H_n(X)/pH_n(X)\simeq H_n(X;\Bbb Z_p)$. Hence, it suffices to show $H_n(X;\Bbb Z_p)\to H_{n-1}(X)$ is injective. How can I?
Edit: My naive approach is that if $\partial_*$ be a connecting homomorphism between $H_n(X;\Bbb Z_p)$ and $H_{n-1}(X)$ and let $\Delta_{*}(X)\otimes\Bbb Z\xrightarrow{i} \Delta_{*}(X)\otimes\Bbb Z\xrightarrow{j}\Delta_*(X)\otimes \Bbb Z_p$ so that $H_n(X)\xrightarrow{i_*}H_n(X)\xrightarrow{j_*} H_n(X;\Bbb Z_p)$, then for $[[c]]\in H_n(X;\Bbb Z_p)$, if $\partial_*([[c]]) = [[i^{-1}\circ (\partial\otimes 1)\circ j^{-1}(c)]] = 0 \iff c=0$ as $i$ maps $\Bbb Z\to p\Bbb Z$. Is this valid?
The point is that your long exact sequence actually takes the form
$\cdots\to H_n(X)\xrightarrow{p} H_n(X)\to H_n(X;\Bbb Z_p)\to H_{n-1}(X)\xrightarrow{p} H_{n-1}(X)\to\cdots$
So $H_n(X;\Bbb Z_p)$ surjects onto its image in $H_{n-1}(X)$, which by exactness is equal to the displayed kernel.
The first isomorphism theorem does not give $H_n(X)/pH_n(X) \cong H_n(X; \Bbb Z_p)$, but rather $H_n(X)/pH_n(X) \cong \operatorname{im}(H_n(X) \to H_n(X;\Bbb Z_p))$, and this image of course injects into $H_n(X;\Bbb Z_p)$ and is also the kernel of the map from $H_n(X;\Bbb Z_p)$ to $H_{n-1}(X)$.