What if $NP\subseteq BPP$? I'm new to complexity and came upon the following exercise which I'm unable to solve.

Prove that if $NP\subseteq BPP$ then $\Sigma_2^p=\Pi_4 ^p$.
Use the Sipser-Lautemann theorem: $\mathrm{BPP} \subseteq \Sigma_2^p \cap \Pi_2^p$.