If $C_\ast$ is a chain complex of vector spaces over any field and $H_\ast$ is its homology, then, viewing $H_\ast$ as a chain complex, there is a chain map $r:C_\ast\to H_\ast$ inducing the obvious isomorphism in homology and there is a chain map $i:H_\ast\to C_\ast$ inducing its inverse. This results in an equivariant splitting $H_\ast^{\otimes n}\to C_\ast^{\otimes n}\to H_\ast^{\otimes n}$. The other factor contributes nothing to the homology of $W\otimes_{\pi}-$, because its homology is trivial and $W$ is free.

To see that the resulting isomorphism between $H_\ast(W\otimes_{\pi} C_\ast^{\otimes p})$ and $H_\ast(W\otimes_{\pi} H_\ast^{\otimes p})$ is independent of the chosen splitting $(r,i)$, first note that it depends only on $i$ and that its inverse (therefore also it) depends only on $r$. Now if $(r',i')$ is another such splitting then the splitting $(r',i)$ yields the same isomorphism as $(r,i)$ and also the same isomorphism as $(r',i')$.

By similar reasoning the isomorphism is natural (with respect to chain maps).

Note that the field did not have to be $\mathbb F_p$, and $p$ did not have to be prime, and the chain complex did not have to come from a space.