= Solution
\b[Charlie cannot send a <bit> to Alice with the stated preparation, fixed interaction and final local measurements.] Write $V=\operatorname{CNOT}_{B\to A}$ and $W=\operatorname{CNOT}_{B\to C}$, so the complete gate is $U=WV$ with the rightmost gate acting first. Alice has finished interacting before the gate involving Charlie acts.
For arbitrary product input <density matrices>, put $\tau_{AB}=V(\rho_A\otimes\rho_B)V^\dagger$. The final <reduced density matrix> of Alice is
$$
\rho_A'=\operatorname{Tr}_{BC}\!\left[W(\tau_{AB}\otimes\rho_C)W^\dagger\right]
=\operatorname{Tr}_{B}\tau_{AB}.
$$
The equality follows because a <unitary operator> on the traced-out systems leaves the <partial trace> unchanged. The right-hand side is independent of Charlie's preparation. Therefore every outcome <probability> of every local measurement by Alice is independent of Charlie's message. <Quantum no-signalling> applies after Alice's last interaction, even though <phase kickback> can allow communication between the two <qubits> of an individual <CNOT gate>.
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