Splitting of the short exact sequence $0\to \frac{H_n(X)}{p H_n(X)}\to H_n(X,\mathbb{Z}_p)\to \ker\{ p: H_{n-1}(X)\to H_{n-1}(X)\}\to 0$

One of the assignments for my algebraic topology course is to solve the following problem from Topology and Geometry by Bredon.

Multiplication by the prime $$p$$ fits a short exact sequence $$0\to \mathbb{Z}\stackrel{p}{\to}\mathbb{Z}\to \mathbb{Z}_p\to 0.$$ Use this to derive the natural split exact sequence $$0\to \frac{H_n(X)}{p H_n(X)}\to H_n(X; \mathbb{Z}_p) \to \ker\{ p: H_{n-1}(X)\to H_{n-1}(X)\}\to 0$$ (The splitting is not natural).

Where $$H_n(X; \mathbb{Z}_p)$$ is the homology of the chain complex $$\Delta_*(X)\otimes \mathbb{Z}_p$$.

I have derived the short exact sequence as follows. A short exact sequence of the form given induces a short exact sequence of chain complexes: $$0\to \Delta_*(X)\otimes \mathbb{Z}\stackrel{1\otimes p}\to \Delta_*(X)\otimes \mathbb{Z}\stackrel{1\otimes \pi}\to \Delta_*(X)\otimes \mathbb{Z}_p\to 0$$ Using the natural isomorphism $$\Delta_*(X)\otimes \mathbb{Z}\cong \Delta_*(X)$$ we have the following short exact sequence of chain complexes:$$0\to \Delta_*(X)\stackrel{p}\to\Delta_*(X)\stackrel{\pi}\to \Delta_*(X)\otimes \mathbb{Z}_p\to 0$$ We recall that such a short exact sequence induces a long exact sequence of homology groups: $$\cdots\to H_{n}(X)\stackrel{p_*}{\to} H_n(X)\stackrel{\pi_*}{\to} H_n(X; \mathbb{Z}_p)\stackrel{\partial_*}{\to} H_{n-1}(X)\stackrel{p_*}{\to} H_{n-1}(X)\to \cdots$$ With $$\partial_*[[c]]:= [p_*^{-1}\circ \partial\circ \pi^{-1}(c)]]$$ for all $$c\in Z_n(\Delta_*(X)\otimes \mathbb{Z}_p)$$. We can construct the following sequence $$0\to \mathrm{im}(\pi_*)\stackrel{\iota}{\to}H_n(X;\mathbb{Z_p})\stackrel{\partial_*}{\to}\ker \{p:H_{n-1}(X)\to H_{n-1}(X)\}\to 0$$ and $$\textrm{im}(\pi_*)\cong \frac{H_n(X)}{\ker \pi_*}=\frac{H_n(X)}{\textrm{im}(p_*)}=\frac{H_n(X)}{pH_n(X)}$$ by the first isomorphism theorem and exactness. Exactness of this sequence follows from the exactness of the long exact sequence. This gives us the desired short exact sequence $$0\to \frac{H_n(X)}{pH_n(X)}\to H_n(X;\mathbb{Z}_p)\to \ker\{p: H_n(X)\to H_n(X)\}\to 0$$

I have derived the short exact sequence, but how can I find a splitting? Is there any intuition that might lead me in the right direction?

Your derivation of the short exact sequence is correct. The cop-out answer to why the sequence is split is to note that it is actually not just a sequence of $$\mathbb{Z}$$-modules, but a sequence of $$\mathbb{Z}_p$$-modules, which is split by linear algebra, because $$\mathbb{Z}_p$$ is a field. This might be useful to consider, but it's not quite enlightening, so let's try constructing an explicit splitting.
I don't think there's too much of an intuition besides simply wanting to "reverse" the diagram chase we do in order to construct the boundary map $$H_n(X;\mathbb{Z}_p)\rightarrow H_{n-1}(X)[p]$$ (this notation just means $$p$$-torsion, but it's shorter). So how does this map work? A homology class in $$H_n(X;\mathbb{Z}_p)$$ is represented by a cycle in $$\Delta_n(X)\otimes\mathbb{Z}_p$$, which we can lift to a chain in $$\Delta_n(X)$$. The differential of this chain is a chain in $$\Delta_{n-1}(X)$$, which is divisible by $$p$$, and dividing it by $$p$$ yields a cycle representing the desired homology class in $$H_{n-1}(X)[p]$$. So, what can go wrong if we try to do it the other way round? Start with a homology class $$[\alpha]\in H_{n-1}(X)[p]$$. This means $$[p\alpha]=p[\alpha]=0$$, so we can find a chain $$c\in\Delta_n(X)$$, such that $$\partial_nc=p\alpha$$. Then, we can simply take $$c\mod p$$ and this defines a homology class in $$H_n(X;\mathbb{Z}_p)$$, because $$\partial(c\mod p)=\partial(c)\mod p=p\alpha\mod p=0$$. You can check that $$[c\mod p]$$ is independent of the choice of representative $$\alpha\in Z_{n-1}(X)$$, but it does depend on the choice of $$c$$.
The solution to this conundrum is that we can find a right-inverse to the differential $$\partial_n\colon\partial_n^{-1}(p\Delta_{n-1}(X))\rightarrow p\Delta_{n-1}(X)\cap B_{n-1}(X)$$ (since the codomain is free), call it $$s\colon p\Delta_{n-1}(X)\cap B_{n-1}(X)\rightarrow\partial_n^{-1}(p\Delta_{n-1}(X))$$ (note that domain and codomain are precisely the two submodules we passed through in our incomplete sketch). Now, I leave it to you to check that $$H_{n-1}(X)[p]\rightarrow H_n(X;\mathbb{Z}_p),\,[\alpha]\mapsto[s(p\alpha)\mod p]$$ is well-defined and a right-inverse to the differential $$H_n(X;\mathbb{Z}_p)\rightarrow H_{n-1}(X)[p]$$ (note that it's a right-inverse, but usually not a left-inverse, precisely because $$s$$ is right-inverse to $$\partial_n$$, but not left-inverse).