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Since $P \subseteq NP$, $NP^P = NP$. Nondeterministic programs are an extension of deterministic programs. They are just programs with a jump-and-branch command added. So every deterministic program is also a nondeterminnistic program that doesn't use jump-and-branch, so adding the oracle gives no additional computability.


This is a good example of how bad notation can make simple questions much more difficult. (1) There is no reason to introduce grammars. You may as well denote by $L_1$ the language accepted by $M_1$ and by $L_2$ the language accepted by $M_2$. (2) Let us denote by $L^c$ the complement of a language $L$. Then your first (correct) equality amounts to prove ...

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