= Solution
To use $S$ as the apparatus, prepare it in the ready state \b[$|+\rangle_S$] and later read it in the <Hadamard basis>. For an arbitrary input $\alpha|+\rangle_D+\beta|-\rangle_D$, part (b) gives
$$
|+\rangle_S(\alpha|+\rangle_D+\beta|-\rangle_D)
\longmapsto
\alpha|+\rangle_S|+\rangle_D+\beta|-\rangle_S|-\rangle_D.
$$
The two apparatus states are orthogonal, so its readout measures the corresponding projectors on $D$ and preserves each input <eigenstate>. The observable is
$$
\boxed{X_D=\widetilde Z_D,}
$$
with outcomes $+1$ and $-1$. This is the <measurement direction of a CNOT interaction> in the complementary basis.
The ready state and readout basis matter. Keeping $S$ in the original $|0\rangle_S$ would make the old-basis controlled gate act as the identity on $D$, producing no record. Thus the same interaction can serve either measurement direction, with different apparatus preparation and pointer readout; the <Hamiltonian operator> alone does not assign an intrinsic system/apparatus role.
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