$\newcommand\ep\varepsilon$Let $X:=\Omega$. Suppose that $|B_x(\ep)\cap X|>0$ for all $x\in X$ and all real $\ep>0$, where $|\cdot|$ is the Lebesgue measure and $B_x(\ep)$ is the open ball in $\mathbb R^m$ of radius $\ep$ centered at $x$. Then the condition $p(x)>0$ for all $x\in X$ implies that $P(p_1\in B_x(\ep)\cap X)>0$ and hence $P(p_1\notin B_x(\ep)\cap X)<1$, for all $x\in X$ and all real $\ep>0$.
On the other hand, by the compactness of $X$, for each real $\ep>0$ there is a finite set $F_\ep\subseteq X$ such that $X\subseteq\bigcup_{x\in F_\ep}B_x(\ep)$.
So, for each real $\ep>0$, $$P(\zeta_n\ge2\ep)\le P(\gamma_n\ge2\ep)\le P\Big(\bigcup_{x\in F_\ep}\bigcap_{i=1}^k\{p_i\notin B_x(\ep)\}\Big) \le\sum_{x\in F_\ep}P(p_1\notin B_x(\ep))^n\to0$$ as $n\to\infty$.
Thus, $\gamma_n\to0$ and $\zeta_n\to0$ in probability, and hence in $L^q$ for all real $q>0$, since $\zeta_n\le\gamma_n\le D<\infty$, where $D$ is the diameter of the compact set $X$.
It should be straightforward to quantify this qualitative argument, in terms of Dudley's entropy number and explicitly given lower bounds on $|B_x(\ep)\cap X|$ and density $p(x)$.