= Solution
Expanding either <Bell state> and using $\langle0|1\rangle_B=0$ removes the cross terms:
$$
\begin{aligned}
\langle\Phi^\pm|Q\otimes I|\Phi^\pm\rangle
&=\frac12(\langle00|\pm\langle11|)(Q\otimes I)(|00\rangle\pm|11\rangle)\\
&=\frac12\bigl(\langle0|Q|0\rangle+\langle1|Q|1\rangle\bigr)\\
&=\frac12\operatorname{tr}Q.
\end{aligned}
$$
\b[Thus the two expectation values agree. Equivalently, both states have the same <reduced density matrix> $I/2$ on qubit $A$; the positivity of $Q$ makes it a possible measurement effect but is not needed for the algebraic equality.]
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