Suppose that $h: \mathbb{Z}\rightarrow \mathbb{R}$ is such that $\mathbb{E}[|h(Z)|]< \infty$.
I have to prove that $$\mathbb{E}[h(X)\mid X^2]=h(|X|)\dfrac{p(|X|)}{p(|X|)+p(-|X|)}+h(-|X|)\dfrac{p(-|X|)}{p(|X|)+p(-|X|)}.$$
I know that $$\psi(y)=\sum_x h(x)P(X=x\mid X^2=y)=\dfrac{\sum_x h(x)P(X=x \land X^2=y)}{P(X^2=y)}$$ if $y > 0$, $\sum_x h(x)P(X=x \land X^2=y)$ has two expressions, because $x=\pm\sqrt y$.
How to write the solution correctly?
Presumably $X$ takes values on $\mathbb{Z}$. Then, $X^2$ takes values on $\{n^2:n\in\mathbb{Z}\}\subset\mathbb{Z}_+=\{0\}\cup\mathbb{N}$. $\sigma(X^2)$ is generated by the sets $A_n=\{\omega: X(\omega)\in\{-n,n\}\}$, $n\in\mathbb{Z}_+$. Hence $$E[h(X)|X^2](\omega)=\sum_{n\geq0}\alpha_n\mathbb{1}_{A_n}(\omega)=\sum_{n\geq0}\alpha_n\mathbb{1}_{\{n^2\}}(X^2(\omega))=:H(X^2(\omega))$$ where $H(t)=\sum_{n\geq0}\alpha_n\mathbb{1}_{\{n^2\}}(t)$, and $\alpha_n\in\mathbb{R}$ are constant to be determined.
For each $n\in\mathbb{Z}_+$, $$E[h(X)\mathbb{1}_{\{n^2\}}(X^2)]=P[X=-n]h(-n)+P[X=n]h(n)$$ On the other hand, $$E\Big[H(X^2)\mathbb{1}_{\{n^2\}}(X^2)\Big]=\alpha_n\big(P[X=-n]+P[X=n]\big)$$ whence we obtain that $$\alpha_n=\frac{P[X=-n]h(-n)+P[X=n]h(n)}{P[X=-n]+P[X=n]}$$ with the understanding that $\alpha_n=0$ if $P[X\in\{-n,n\}]=0$.
Putting things together $$E[h(X)|X^2]=\sum_{n\geq0}\frac{P[X=-n]h(-n)+P[X=n]h(n)}{P[X=-n]+P[X=n]}\mathbb{1}_{\{n^2\}}(X^2)$$
