= Solution
Normalization requires $2p+2q=1$, so $q=1/2-p$ and $0\leq p\leq1/2$. With logarithms to base two, the <Shannon entropy> is
$$
\boxed{H(X)=-2p\log_2p-2q\log_2q=1+h_2(2p).}
$$
The convention $0\log0=0$ gives $H=1$ at both endpoints. In the interior,
$$
H'(p)=2\log_2\frac qp,\qquad
H''(p)=-\frac2{\ln2}\left(\frac1p+\frac1q\right)<0.
$$
Hence the unique maximum is $H=2$ bits at $p=q=1/4$. The curve is concave and symmetric about $p=1/4$.
Choose the message assignment $x_1\leftrightarrow\Psi^+$, $x_2\leftrightarrow\Psi^-$, $x_3\leftrightarrow\Phi^+$ and $x_4\leftrightarrow\Phi^-$. Starting from $|\Psi^+\rangle$, apply respectively $I,Z,X,XZ$ to the first <qubit>. This produces those four orthonormal <Bell states>, up to a physically irrelevant overall sign in the last case. The ensemble <density operator> is
$$
\boxed{\rho_Y=p\Pi_{\Psi^+}+p\Pi_{\Psi^-}+q\Pi_{\Phi^+}+q\Pi_{\Phi^-}
=\operatorname{diag}(q,p,p,q)}
$$
in computational order $00,01,10,11$, where $\Pi_B=|B\rangle\langle B|$. The cancellation of off-diagonal terms follows from the equal probabilities within each sign pair; the individual encoded states remain entangled.
A <Bell basis> measurement reads the label perfectly because the four code states are orthogonal. Thus this is equivalent to a classical error-free encoding when unrestricted joint measurement is allowed. Each conditional encoded state is pure, so
$$
\boxed{H(Y|X)=\sum_i p_i S(\Pi_i)=0,\qquad
H(X:Y)=S(\rho_Y)-\sum_i p_iS(\Pi_i)=H(X).}
$$
The second formula is the classical-quantum <mutual information>, here equal to attainable classical information. The first says there is no uncertainty about the joint pure code state once the message is known; it does not say that measuring its individual <qubits> has deterministic outcomes.
\Image[/past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2015-message-entropy-and-zeno-transfer.png]
{title=Message <Shannon entropy> and measurement-guided state transfer}
{description=Original theoretical curves: $H(X)=1+h_2(2p)$ for the specified four-message distribution, and finite-step Zeno transfer probabilities for a known initial basis state. The conditioned all-following probability and the unconditioned final-target probability differ at finite step count but both tend to one.}
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