= Solution
With $\beta=1/(k_BT)$, the <canonical partition function> and equilibrium probabilities are
$$
\boxed{Z=\sum_ne^{-\beta E_n},\qquad p(n)=\frac{e^{-\beta E_n}}Z}.
$$
Differentiating $T\log Z$, with fixed energy levels, gives $S=k_B\log Z+\langle E\rangle/T$. Since $\log p(n)=-\beta E_n-\log Z$, normalization yields
$$
\boxed{S=-k_B\sum_np(n)\log p(n)}.
$$
The sums include degeneracy by labeling states rather than distinct energy values, and a zero probability contributes zero by continuity.
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