= Solution
Alice makes a two-outcome local <quantum measurement> with <Kraus operators>
$$
K_{\rm s}=\begin{pmatrix}\tan\alpha/\tan\beta&0\\0&1\end{pmatrix},\qquad
K_{\rm f}=\begin{pmatrix}\sqrt{1-\tan^2\alpha/\tan^2\beta}&0\\0&0\end{pmatrix}.
$$
Since $0<\tan\alpha/\tan\beta<1$, these are valid and satisfy $K_{\rm s}^\dagger K_{\rm s}+K_{\rm f}^\dagger K_{\rm f}=I$. Alice sends Bob the success or failure outcome. The successful unnormalized branch is
$$
\begin{aligned}
(K_{\rm s}\otimes I)|\psi\rangle
&=\frac{\sin\alpha\cos\beta}{\sin\beta}|00\rangle+\sin\alpha|11\rangle\\
&=\frac{\sin\alpha}{\sin\beta}|\phi\rangle.
\end{aligned}
$$
Therefore the normalized success state is exactly the target and
$$
\boxed{p_{\rm succ}=\frac{\sin^2\alpha}{\sin^2\beta}=\frac{s}{t}=p_{\max}.}
$$
Bob needs no quantum correction in this known Schmidt basis. The failure branch is proportional to $|00\rangle$, a <product state>, and has probability $1-s/t$. The successful branch saturates the previous bound: $p_{\rm succ}E_2(\phi)=E_2(\psi)$ and failure contributes zero. This establishes <optimal stochastic conversion of a two-qubit pure state>.
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