Implications of $NP = \Sigma_2 P$ for PH collapse
The answer is yes.
notice that $NP=\Sigma_1$ and for all $i$ you have $\Sigma_i=\Sigma_{i+1} \Rightarrow \Sigma_i=\Pi_i$. This is true since $\Pi_i\subseteq \Sigma_{i+1}=\Sigma_i$. The inclusion holds since if $\bar{L}\in\Sigma_i$ then you have $x\notin L \iff \exists v_1\in \left\{0,1\right\}^{p(|v_1|)}...Q_iv_i\in\left\{0,1\right\}^{p(|v_i|)}M(x,v_1,...,v_i)=1 $ or equivalently $x\in L \iff \forall v_1\in \left\{0,1\right\}^{p(|v_1|)}...\bar{Q_i}v_i\in\left\{0,1\right\}^{p(|v_i|)}M(x,v_1,...,v_i)=0$. So now your $\Sigma_{i+1}$ verifier for $L$ will ignore $v_1$, run $M(v_2,...,v_{i+1})$ and flip the answer. It remains to show $\Sigma_i\subseteq\Pi_i$. Let $L\in\Sigma_i, \bar{L}\in \Sigma_{i+1}$ and since $\Sigma_i=\Sigma_{i+1}$ you get $\bar{L}\in\Sigma_i$.

Back to your question. If $NP=\Sigma_2$ (or $\Sigma_1=\Sigma_2)$ then $\Sigma_1=\Pi_1$. Let $L\in\Pi_2$ then $\bar{L}\in\Sigma_2\Rightarrow \bar{L}\in\Sigma_1\Rightarrow \bar{L}\in\Pi_1\Rightarrow L\in\Sigma_1$, So $L\in NP$ as required. The opposite direction is clear(for all $i$ $\Sigma_i\subseteq\Pi_{i+1}$).