= Solution
All time-dependent <Hamiltonian operators> are multiples of the same fixed operator, so they commute and their <time-ordered exponential> reduces to an ordinary exponential. Let $P_1=(I-Z_S)/2$. With $\hbar=1$ and the given unit coupling integral,
$$
U=\exp\left[-\frac{i\pi}{2}P_1\otimes(I-X_D)\right]
=P_0\otimes I+P_1\otimes e^{-i\pi(I-X_D)/2}.
$$
Since $X_D^2=I$,
$$
e^{-i\pi(I-X_D)/2}
=e^{-i\pi/2}\left[\cos(\pi/2)I+i\sin(\pi/2)X_D\right]=X_D.
$$
Thus
$$
\boxed{U=P_0\otimes I+P_1\otimes X_D,\qquad
U|s,d\rangle=|s,d\oplus s\rangle.}
$$
It is exactly the <CNOT gate> with $S$ controlling $D$, with no residual relative phase. Acting on $(c_0|0\rangle+c_1|1\rangle)_S|0\rangle_D$ correlates the system basis label with the apparatus label, producing $c_0|00\rangle+c_1|11\rangle$ as required.
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