= Solution
Encode the ensemble label in an orthogonal classical register:
$$
\omega_{XB}=\sum_xp_x|x\rangle\langle x|\otimes\rho_x.
$$
The <classical-quantum state> has $S(XB)=H(p)+\sum_xp_xS(\rho_x)$, $S(X)=H(p)$ and $S(B)=S(\bar\rho)$, where $\bar\rho=\sum_xp_x\rho_x$. Hence
$$
I(X:B)_\omega=S(\bar\rho)-\sum_xp_xS(\rho_x)=\chi(\mathcal E).
$$
Applying $\operatorname{id}_X\otimes\Lambda$ changes the conditional states to the output ensemble. The same identity gives $I(X:B')=\chi(\mathcal E')$. <Data processing for quantum mutual information>, proved in Question 3, therefore yields the <Holevo quantity under a quantum channel>:
$$
\boxed{\chi(\mathcal E')\leq\chi(\mathcal E).}
$$
This concerns the information available in the whole ensemble, not the change in the entropy of each individual state.
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