If a language consists of an NP and coNP question, do we have to place it in P^NP^NP?
No, because $A$ isn't necessarily $NP$-complete and $B$ isn't necessarily $coNP$-complete.  You'd need both to be true to force the implication $NP = coNP$ by $X$ being a member of both problem classes.  In fact $A$ and $B$ could both be in $NP$ and $coNP$.  E.g. $A$ and $B$ could both be in $P$.  $NP \cap coNP$ is not the empty set.