= Solution
{c}
Let $S$ denote immediate transmission through the BS1 subsystem between Bob's lower-right port and Alice's upper-left port; let $L$ denote transmission after one visit to each internal mirror. From the same <beam splitter> rules as in <6/b>, the corresponding amplitudes in either direction are
$$
S=c,\qquad L=-c^3
$$
up to a common fixed two-mirror propagation phase. The minus sign in $L$ comes from the $1'\to1$ or reverse $1\to1'$ scattering. The two paths selected by D0's time gate are $L$ on the outward journey followed by $S$ on the return, and $S$ outward followed by $L$ on the return. They have the same total delay and fixed propagation phase, so their <probability amplitudes> interfere.
For the first history, Alice applies $e^{i\phi_A}$ to the late component, while Bob's return-time modulation is not encountered in the return delay arm. For the second, Alice's early component is unmodulated and Bob applies $e^{i\phi_B}$ in the return delay arm. Bob must time his modulation to the return passage, leaving the outward reference preparation unchanged; this is implicit in the stated encoding. Each enabled modulator is traversed twice, so its $\pi/2$ single-pass <phase> becomes the stipulated $\pi$.
Include the amplitude $c$ to enter BS1 through BS2, and the amplitude of modulus $c$ to reach D0 through BS2 on return. Both histories therefore contain six 50:50 <beam splitter> factors. Ignoring one common unobservable <global phase>,
$$
\mathcal A_{D0}=-c^6(e^{i\phi_A}+e^{i\phi_B})
=-\frac18(e^{i\phi_A}+e^{i\phi_B}).
$$
Hence the \b[unconditional probability per emitted <photon>], with no renormalization after discarding paths, is
$$
\boxed{P(D0\mid\phi_A,\phi_B)
=\frac{1+\cos(\phi_A-\phi_B)}{32}
=\frac1{16}\cos^2\!\frac{\phi_A-\phi_B}{2}.}
$$
For the allowed values $0,\pi$, this is $1/16$ when $\phi_A=\phi_B$ and zero otherwise. The ignored trajectories exit other ports or other time bins; excluding them from D0's acceptance does not increase the success probability. Equal mirror reflection phases, channel phases and Alice's mirror phase multiply both accepted histories equally and cancel from the <interference> condition. A residual \i[relative] phase would need calibration.
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