 Solution

Charlie cannot send a bit to Alice with the stated preparation, fixed interaction and final local measurements. Write $V={\mathrm{CNOT}}_{B\to A}$ and $W={\mathrm{CNOT}}_{B\to C}$, 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 ${\tau }_{AB}=V({\rho }_A\otimes {\rho }_B)V^{†}$. The final reduced density matrix of Alice isThe 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.