First of all, let us assume that $\pi$ is onto to avoid  some unnecessary complications. Then we can find linear coordinates $x=(x_1,\dotsc, x_n)$ on $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$ and $y=(y_1,\dotsc, y_m)$  on $\bR^m$ such that, in these coordinates,    $\pi$ is given by

$$ y_j=x_j,.\;\;\forall j=1,\dotsc, m. $$

The probability measure $\newcommand{\bP}{\mathbb{P}}$ $\bP$  then has the form
$$ \bP(dx)= \rho(x) dx. $$

Then  

$$\bP_Y(dy)= \rho_Y(y) dy,\;\;\;\rho_Y(y_1,\dotsc,y_m)=\int \rho(y_1,\dotsc, y_m, x_{m+1}, \dotsc x_n) dx_{m+1}\cdots dx_n.  $$ 

If  $x$ and $y$  are arbitrary *orthonormal* coordinates on $\bR^n$  and resp. $\bR^m$, then $\pi$ is represented  by  a matrix $A$ and we set

$$J_\pi:=\sqrt{\det  AA^*}. $$

This quantity , called the } of $\pi$ is *independent* of  the choice of orthonormal coordinates $x$ and $y$. 

If $B\subset \bR^m$ is a Borel set, then the coarea formula 

$$ \int_{\pi^{-1}(B)} \rho(x) dx = \int_B \;\underbrace{\left(\int_{\pi^{-1}(y)}\frac{\rho(x)}{J_\pi} dV_{\pi^{-1}(y)} (x)\right)}_{=:\rho_Y(y)}\; dy, $$  

where $dV_{\pi^{-1}(y)}$ denotes the volume element on the affine  plane $\pi^{-1}(y)$ induced by the     Euclidean metric on $\bR^n$.   For more details check the first two pages of [these notes][1].


  [1]: http://www.nd.edu/~lnicolae/Coarea.pdf