= Solution
Write the normalized input <quantum state> as $|\psi\rangle=a|0\rangle+b|1\rangle$. The fresh <quantum ancilla> is $|+\rangle$, so the <Controlled-Z gate> creates
$$
a|0\rangle|+\rangle+b|1\rangle|-\rangle.
$$
Applying the measurement bra $\langle v_k(\theta)|$ to the first <qubit> leaves the unnormalized second-<qubit> vector
$$
|\widetilde\psi_k\rangle=\frac1{\sqrt2}\left(a|+\rangle+(-1)^k e^{-i\theta}b|-\rangle\right).
$$
Since $|+\rangle,|-\rangle$ form an <orthonormal basis>,
$$
\Pr(k)=\langle\widetilde\psi_k|\widetilde\psi_k\rangle
=\frac{|a|^2+|b|^2}{2}=\frac12.
$$
Also $U(\theta)|\psi\rangle=a|+\rangle+e^{-i\theta}b|-\rangle$, while the <Pauli X gate> fixes $|+\rangle$ and changes the sign of $|-\rangle$. Dividing by the branch norm therefore gives
$$
\boxed{|\psi'_k\rangle=X^kU(\theta)|\psi\rangle,\qquad\Pr(k=0)=\Pr(k=1)=\frac12.}
$$
This is <one-bit teleportation> with the present positive-exponent measurement vectors; the transferred logical gate is $U(\theta)=J(-\theta)$.
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