= Solution
Choose a finite symmetric generating set $S$ for $\Gamma$, let $B_S(m)$ be its <word metric> ball, and fix $p\in M$. Set $D=\max_{s\in S}d(p,sp)$. The <triangle inequality> and invariance of distance imply
$$
d(p,\gamma p)\leq Dm\qquad(\gamma\in B_S(m)).
$$
The <properly discontinuous group action> has a finite <point stabilizer> $H=\Gamma_p$. It also gives uniform separation of distinct orbit points. To see this, only finitely many <group> elements can send $p$ into a fixed <compact> ball about $p$, by proper discontinuity and the <Hopf-Rinow theorem>. The positive distances among this finite list, together with the radius of the ball for all other elements, have a positive lower bound $\delta$. Thus $d(\gamma p,\gamma'p)\geq\delta$ whenever the centers differ.
Take $0<\varepsilon<\delta/2$ and write $v=\operatorname{Vol}B(p,\varepsilon)>0$. Balls of radius $\varepsilon$ about distinct orbit points are disjoint and all have volume $v$, because the action is by <isometries>. Those corresponding to words of length at most $m$ lie in $B(p,Dm+\varepsilon)$. Each orbit point arises from at most $|H|$ <group> elements in $B_S(m)$, so
$$
\frac{|B_S(m)|}{|H|}\,v\leq\operatorname{Vol}B(p,Dm+\varepsilon)
\leq\omega_n(Dm+\varepsilon)^n.
$$
The final inequality is the allowed <Bishop-Gromov inequality> for nonnegative <Ricci curvature>. If the entire <group> fixes $p$, proper discontinuity already makes it finite and the conclusion is immediate. Otherwise the preceding separation argument applies. Therefore
$$
\boxed{|B_S(m)|\leq\frac{|H|\omega_n}{v}(Dm+\varepsilon)^n\leq C(1+m)^n.}
$$
This is <polynomial growth of a group> of degree at most $n$, proved by <polynomial group growth from orbit packing>. Neither freeness of the action nor <compactness> of the quotient was assumed.
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