 Solution

Use basis order $000,001,010,011,100,101,110,111$, with the ancilla first, and put $x=e^{i{\varphi }_1}$, $y=e^{i{\varphi }_2}$. Initially the ancilla is at the north pole of the Bloch sphere. The other two qubits lie on its equator at azimuths ${\varphi }_1,{\varphi }_2$, with Bloch vectors $(\cos {\varphi }_j,\sin {\varphi }_j,0)$.

The four dotted cuts, in left-to-right order, have the explicit normalized vectorsThe first Hadamard gate copies the lower-register amplitudes into two control branches; the Fredkin gate exchanges them only in the second branch; the final Hadamard gate adds and subtracts these branches.

Summing the squared magnitudes of the first four entries of $v_3$ givesMeanwhile $⟨{\varphi }_2|{\varphi }_1⟩=(1+e^{i({\varphi }_1-{\varphi }_2)})/2$, soIdentical input qubits give $P_0=1$; orthogonal equatorial inputs give $P_0=1/2$. The factor one half is an ancilla-interference effect, not a direct projective measurement of the lower qubits individually.