= Solution
The <linear operator> $M_A$ acts on the first factor, while the partial inner product acts on the second. They therefore commute in the contraction:
$$
\begin{aligned}
(I_A\otimes\langle\phi_B^*|)(M_A\otimes I_B)|\widetilde\Psi_{AB}\rangle
&=M_A(I_A\otimes\langle\phi_B^*|)|\widetilde\Psi_{AB}\rangle\\
&=\boxed{M_A|\phi_A\rangle}.
\end{aligned}
$$
Thus the operated vector is exactly the <relative state> obtained from the operated bipartite vector using the same <conjugate index vector>. For a physical normalized <pure state>, divide by $\|M_A\phi_A\|$ when that norm is nonzero. If $M_A\phi_A=0$, the vector identity still holds, but there is no normalized output state for that zero-probability branch. An arbitrary operator is not assumed unitary or norm preserving.
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