That the action is infinitesimally free tells you the orbits are diffeomorphic to $\mathbb R^n / G$ for various discrete subgroups $G$ of $\mathbb R^n$. This is just general nonsense with Lie group actions. And $\mathbb R^n/G$ is diffeomorphic to $\mathbb T^k \times \mathbb R^{n-k}$, that's pretty much just linear algebra -- take a basis for $G$ as a $\mathbb Z$-module, and extend that to a basis for $\mathbb R^n$, that gives you your diffeomorphism. In this argument $k$ need not be constant.
The last thing you're interested in is showing $k$ is constant and that the orbit types are constant. This apparently is the reason for the assumption of the bundle. The main thing this seems to be telling you is that the orbits are closed. Because if the orbit is closed you can construct a transversal and apply the above argument. This tells your your manifold $M$ fibers over something with fiber is $\mathbb T^k \times \mathbb R^{n-k}$, and that something would be $P$.
So I don't think there's a real reference for this result, it's just a standard putting-together of basic results about group actions. Take a look at Conlon's book on manifolds -- he gives some similar (but not quite the same) types of arguments concerning actions, commuting vector fields and such.