Picking up some of Eric Zaslow's reformulation: Assume $P$ is commutative, saturated, and cancellative, as well as finitely generated. (Cancellative means $p_1+q=p_2+q$ implies $p_1=p_2$.) The answer to your question is affirmative if and only if the "groupification" $P^{gp}$ of $P$ is torsion-free. (As mentioned in Dustin's answer, saturated means that for all $p$ in $P^{gp}$, $np \in P$ implies $p\in P$. Cancellative means $p_1+q=p_2+q$ implies $p_1=p_2$.)
All this amounts to $P$ being embeddable as a sub-monoid of ${\Bbb Z}^n$ for some $n$. Then take the subgroup of ${\Bbb Z}^n$ spanned by $P$. This is isomorphic to some ${\Bbb Z}^m$; take the dual of the convex hull of $P$ in ${\Bbb R}^m$ and you've got your cone $\sigma$, just as Eric says. When $P$ is saturated, it is equal to $\sigma^\vee \cap M$; otherwise, this gives the saturation of $P$, corresponding to the integral closure of $k[P]$.
Depending on what references you use, when $P^{gp}$ is torsion-free, $P$ is called either integral or toric. (See, e.g., the toric variety notes on M. Mustata's webpage versus the log geometry notes on Danny Gillam's webpage; both sources are worth looking at.) It seems the latter terminology is more standard in the log geometry world, where "integral" sometimes just means "cancellative".