I will assume that by an "arithmetic group" you mean a subgroup $\Gamma$ commensurable to the intersection $GL(n, {\mathbb Z})\cap G$, where $G< GL(n, {\mathbb R})$ is an algebraic subgroup defined over a number field $F$; actually, $F={\mathbb Q}$ will suffice for my purposes. (I prefer to use Greek letters to denote arithmetic subgroups, as it is customary in the literature.)

I will even assume that $G$ is semisimple (which you are probably implicitly assuming). I claim that, in general, $Q$ does not exist.

The example will be given by $n=3, G=O(2,1)$, the stabilizer of the quadratic form $q=-x_0^2+x_1^2+x_2^2$. I will use $\Gamma<G\cap GL(3, {\mathbb Z})$, the index two subgroup preserving the future light cone $x_0>0$. Take $p=(1,0,0)$ and $P=\{p\}$. Then the orbit $\Gamma \cdot p$ is an unbounded subset of the open future light cone $L_+$. Moreover, the convex hull $C$ of this orbit will contain the domain $$ \{x\in L_+: q(x)\le -1\}. $$ Now, if $Q\subset L_+$ is any bounded subset, then its $G$-orbit (hence, $\Gamma$-orbit) is contained in a region $$ \{x\in L_+: a\le q(x)\le b\} $$ for a pair of real numbers $a, b$ satisfying $b<a<0$. Hence, no bounded subset $Q$ of $L_+$ will satisfy the containment $$ C\subset \Gamma \cdot Q. $$ If $Q$ is not contained in $L_+$, then we of course cannot have the other containment $$ \Gamma \cdot Q\subset C. $$

If yo allow reductive groups $G$, then one can even get an example with $n=2$.

However, maybe by an "arithmetic group" you simply mean a finite index subgroup $\Gamma$ of $GL(n, {\mathbb Z})$. Then one can prove that for every $P$ either $C={\mathbb R}^n$ or $P=C=\{0\}$. In this case, $Q$ of course exists and for $C={\mathbb R}^n$ you can take $Q$ to be any convex polytope whose interior contains the origin.

I will assume that by an "arithmetic group" you mean a subgroup $\Gamma$ commensurable to the intersection $GL(n, {\mathbb Z})\cap G$, where $G< GL(n, {\mathbb R})$ is an algebraic subgroup defined over a number field $F$; actually, $F={\mathbb Q}$ will suffice for my purposes. (I prefer to use Greek letters to denote arithmetic subgroups, as it is customary in the literature.)

I will even assume that $G$ is semisimple (which you are probably implicitly assuming). I claim that, in general, $Q$ does not exist.

The example will be given by $n=3, G=O(2,1)$, the stabilizer of the quadratic form $q=-x_0^2+x_1^2+x_2^2$. I will use $\Gamma<G\cap GL(3, {\mathbb Z})$, the index two subgroup preserving the open future light cone $$ L_+=\{x: q(x)<0, x_0>0\}.$$ Take $p=(1,0,0)$ and $P=\{p\}$. Then the orbit $\Gamma \cdot p$ is an unbounded subset of   $L_+$. Moreover, for each point $x$ in the convex hull $C$ of this orbit, $C$ also contains the cone $x+ L_+$. This is because $\Gamma$ is a lattice in $G$ and, hence, the closure of the projection of $\Gamma x$ to ${\mathbb R} P^2$ contains the quadric $$ P(\{q(x)=0\})\subset {\mathbb R} P^2. $$ In particular, $q$ is unbounded from below on $C$.

Now, if $Q\subset L_+$ is any compact subset, then its $G$-orbit (hence, $\Gamma$-orbit) is contained in a region $$ \{x\in L_+: a\le q(x)\le b\} $$ where $$ a= \max q|_Q, b=\min q|_Q. $$ Hence, no bounded subset $Q$ of $L_+$ will satisfy the containment $$ C\subset \Gamma \cdot Q. $$ If $Q$ is not contained in $L_+$, then we of course cannot have the other containment $$ \Gamma \cdot Q\subset C. $$

If yo allow reductive groups $G$, then one can even get an example with $n=2$.

However, maybe by an "arithmetic group" you simply mean a finite index subgroup $\Gamma$ of $GL(n, {\mathbb Z})$. Then one can prove that for every $P$ either $C={\mathbb R}^n$ or $P=C=\{0\}$. In this case, $Q$ of course exists and for $C={\mathbb R}^n$ you can take $Q$ to be any convex polytope whose interior contains the origin.