Let $\pi:E\to M$ be a vector bundle over a closed smooth manifold and supose $\Pi:F\to E$ is a fibre bundle over the total space of $\pi$. I'd like to know if, restricted to $E_p$, the second bundle $Pi$ is trivializable (moreover, homogeneous) as a fibre bundle over the vector space $E_p$. I belive this is true if $\Pi$ is a vector bundle over $E$, but I haven't written down a proof yet. I'll appreciate any help. Thanks.
If $E \to M$ is a vector bundle then it is a weak equivalence. (being a vector bundle means it is a fibration and then look at the LES in homotopy and as Somnath points out the fiber is contractible.) So having a bundle on E is the same as having a bundle on M up to homotopy.
But as for your question, up to isomorphism there is only one vector bundle over a contractible space. It is definitely trivial over each fiber of the vector bundle you started with.