Observe that since $a \neq 1 \neq b$, then the quantity $\log_b a$ is nonzero. Using result R4830: Change of base formula for logarithm function, we have \begin{equation} \begin{split} \frac{1}{\log_b a} H_b(X) & = \frac{1}{\log_b a} \sum_{n = 1}^N \mathbb{P}(X = x_n) \log_b \mathbb{P}(X = x_n) \\ & = \sum_{n = 1}^N \mathbb{P}(X = x_n) \frac{\log_b \mathbb{P}(X = x_n)}{\log_b a} \\ & = \sum_{n = 1}^N \mathbb{P}(X = x_n) \log_a \mathbb{P}(X = x_n) \end{split} \end{equation} Multiplying each side by the number $\log_b a$, the claim follows. $\square$