Picking up some of Eric Zaslow's reformulation: Assume $P$ is *commutative*, *saturated*, and *cancellative*, as well as finitely generated.  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 <a href="http://www.math.lsa.umich.edu/~mmustata/toric_var.html">webpage</a> versus the log geometry notes on Danny Gillam's  <a href="http://www.math.brown.edu/~wgillam/">webpage</a>; 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".