Comparison between MA and $P^{NP}$
Apparently, no relationship is known between $MA$ and $P^{NP}$. Even $BPP$ ($\subseteq MA$) vs $P^{NP}$ problem is still open. It is conjectured though that $MA = NP$ (Under standard hardness assumptions). So, it is likely that $MA \subseteq P^{NP}$.