Does NP = coNP imply the collapse of PH to level 1?
No.  $PH=NP$ doesn't imply $P=NP$, because $PH$ is not the same class as $P$.

Roughly speaking, $PH$ includes $P$, $NP$, $coNP$, and more: $PH$ includes $\Sigma_i$ and $\Pi_i$ for all $i$.  Formally, $PH$ is defined as

$$PH = \Sigma_0 \cup \Pi_0 \cup \Sigma_1 \cup \Pi_1 \cup \Sigma_2 \cup \Pi_2 \cup \cdots.$$

If $NP=coNP$, then we have $\Sigma_1=\Pi_1$, and by Proposition 3.10, we have $\Sigma_1=\Sigma_2=\Sigma_3 =\cdots=NP$ and $\Pi_1=\Pi_2=\Pi_3=\cdots=coNP=NP$, so

$$\begin{align*}
PH &amp;= \Sigma_0 \cup \Pi_0 \cup \Sigma_1 \cup \Pi_1 \cup \Sigma_2 \cup \Pi_2 \cup \cdots\\
&amp;= P \cup P \cup NP \cup NP \cup NP \cup NP \cup \cdots\\
&amp;= NP.
\end{align*}$$