Prove that $ \text{H}^1(G/P,\text{Z}(P)) = 0 $ for a normal Sylow p-subgroup P

Question

Let $ p $ be a prime and $ G $ a (multiplicatively written) finite group with a normal Sylow $ p $-subgroup. Let $ Q = G/P $. Then $ Q $ acts on $ \text{Z}(P) $ via conjugation: $ gP.z = gzg^{-1} $.
With this group operation $ \text{Z}(P) $ becomes a $ Q $-module. Now my question is, how can I prove, that $ \text{H}^1(Q,\text{Z}(P) = 0 $?

Clearly $ Q $ is a $ p' $-group. I tried to show that any $ \mathbb{Z}Q $-module homomorphism from $ \mathbb{Z}Q $ to $ \text{Z}(P) $ is trivial but failed until now.