= Solution
Without Alice's operation, Bob obtains outcome $b$ with probability
$$
p(b)=\langle\psi|I_A\otimes|b\rangle\langle b|\psi\rangle.
$$
If Alice applies $U$ and then measures, but her outcome is not communicated, Bob's probability is
$$
\begin{aligned}
p'(b)
&=\sum_a\langle\psi|
U^\dagger\Pi_aU\otimes|b\rangle\langle b|\psi\rangle\\
&=\langle\psi|I_A\otimes|b\rangle\langle b|\psi\rangle=p(b),
\end{aligned}
$$
because $\sum_a\Pi_a=I_A$. Equivalently, a trace-preserving local operation leaves Bob's <reduced density matrix> unchanged. This is the <quantum no-signalling>[no-communication theorem]: local operations cannot signal without communication of their outcomes.
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