First of all, the assertion is not quite true if $X_P$ and $X_Q$ are not embeddings of the torus $T=\mathbf G_m^n$. Let, e.g., $n=3$, $P=conv(0,e_1,e_2,e_3)$, $Q=conv(0,e_1,e_2,e_1+e_2+2e_3)$. Then $N+1=4$ and $X_P=X_Q=\mathbf P^3$.
So assume that $X_P$ and $X_Q$ are embeddings. This means that the lattice $\mathbb Z^n$ is generated by the differences of lattice points. . Then the corresponding homomorphisms $p,q:T\to PGL(N+1)$ are injective. Let $G=\{g\in PGL(N+1)\mid g(X_P)=X_P\}$. Then $G\to Aut(X_P)$ is injective since $X_P$ spans $\mathbb P^N$. Thus $p(T)$ is a maximal torus of $G$. Indeed $T$ is already a maximal torus in the entire Cremona group (see Demazure's paper).
Now assume that $X_P$ and $X_Q$ are projectively equivalent. Then there is $a\in PGL(N+1)$ with $a(X_Q)=X_P$. Then, we get two injective homomorphisms $T\to PGL(N+1)$ namely $p$ and $\bar q:=aqa^{-1}$. Since both $p(T)$ and $\bar q(T)$ are maximal tori of $G$, there is $b\in G$ with $p(T)=b\bar q(T)b^{-1}=cq(T)c^{-1}$ with $c=ba$. It follows that there is an automorphism $\alpha$ of $T$ with $p(\alpha(t))=cq(t)c^{-1}$. Now replace $P$ by $\alpha(P)$. Then we may assume that $\alpha=id_T$. Then $p$ and $q$ are conjugated by an element $c\in PGL(N+1)$.
Finally observe that the projective representation $p:T\to PGL(N+1)$ uniquely determines the weights of $T$ on $\mathbb C^{N+1}$ up to a translation. This set of weights is nothing else than the lattice points in $P$. So the lattice points of $P$ and $Q$, and therefore $P$ and $Q$ themselves, differ by a translation.
