= Solution
Introduce an orthonormal classical label register $X$ and the <classical-quantum state>
$$
\omega_{XQ}=\sum_i p_i|i\rangle\langle i|_X\otimes\rho_i.
$$
The blocks have orthogonal supports because their label vectors are orthogonal, regardless of whether the original $\rho_i$ have orthogonal supports. Part (a) gives $S(XQ)=H(p)+\sum_i p_iS(\rho_i)$. The marginals are $\omega_X=\sum_i p_i|i\rangle\langle i|$ and $\omega_Q=\sum_i p_i\rho_i$, so $S(X)=H(p)$. Apply the <Subadditivity of Von Neumann entropy> and cancel $H(p)$:
$$
H(p)+\sum_i p_iS(\rho_i)\leq H(p)+S\!\left(\sum_i p_i\rho_i\right).
$$
Thus \b[$S(\sum_i p_i\rho_i)\geq\sum_i p_iS(\rho_i)$]. This is the <Concavity of Von Neumann entropy>.
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