Triangulate the Voronoi cell then write the integral as a sum over the triangles:

$$\int_{\Omega}\|P - Pi\|\,dP = \sum_{k=1}^{N}\int_{\Delta_k}\|P-P_i\|\,dP.$$

Write the integration over the triangle in barycebtric coordinates. Let the Jacobian of the transformation for triangle $k$ be $J_k$, and $|det(J_k)| = 2|Area_{\Delta_k}|$. Then $\int_{\Delta_k}f(P)\,dP = |det(J_k)|\int_0^1\int_0^{1-\beta}f(P(\alpha, \beta))\,d\beta\,d\alpha$, where $\alpha, \beta$ are two of the barycentric coordinates.

Now we need only evaluate (let $\vec{v}_i$ be the vertices of the current triangle):
$$\int_0^1\int_0^{1-\beta}\|\vec{v}_1 + \alpha (\vec{v}_2 - \vec{v}_1) + \beta (\vec{v}_3 - \vec{v}_1) - P_i\|^2\,d\alpha\,d\beta$$

Expand this and integrate the polynomials. The Jacobian determinants is: $$|det(J)| = \|(v_2-v_1) \times (v_3 - v_1)\|$$.
If you are working in 3D, $\times$ is the cross product here. If it's 2D then augment the vectors with a $0$ for $Z$ and perform the cross product.