$\newcommand\P{\mathcal P}\newcommand\X{\mathcal X}$A counterexample is as follows: $n=m=1$, $\X=[-1,1]$, $\P=[0,2]$, $f(x)=\sqrt2-1/\sqrt2+|x|^{3/2}$, $g(x)=1/\sqrt2$, and $h(x)=1=S(x)$ for all $x\in\X$.

$\newcommand\P{\mathcal P}\newcommand\X{\mathcal X}$A counterexample is as follows: $n=m=1$, $\X=[-1,1]$, $\P=[0,2]$, $f(x)=\sqrt2-1/\sqrt2+|x|^{3/2}$, $g(x)=1/\sqrt2$, and $h(x)=1=S(x)$ (for all $x\in\X$).

Indeed, then $$\Pi(x)=\{p\in[0,2]\colon d(p)\le|x|^{3/2}\},$$ where $$d(p):=\sqrt{1+p^2}-(\sqrt2-1/\sqrt2+p/\sqrt2),$$ so that $d$ is a convex function with $d(1)=0=d'(1)$ and $d''(1)=2^{-3/2}$. So, $d(p)\sim2^{-5/2}(p-1)^2$ for $p\to1$ and hence $$\Pi(x)=[p_-(x),p_+(x)]$$ for some functions $p_\pm$ such that $p_\pm(x)-1\sim2^{-5/4}|x|^{3/4}$ as $x\to0$. Since $3/4<1$, the function $\Pi$ is not Lipschitz in any neighborhood of $0$.