Using results
\begin{equation}
\begin{split}
H_a(X \mid X)
& = - \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} \mathbb{P}(X = x, X = y) \log_a \mathbb{P}(X = x \mid X = y) \\
& = - \sum_{x \in \mathcal{X}} \mathbb{P}(X = x, X = x) \log_a \mathbb{P}(X = x \mid X = x) \\
& = - \sum_{x \in \mathcal{X}} \mathbb{P}(X = x) \log_a 1 \\
& = - \sum_{x \in \mathcal{X}} \mathbb{P}(X = x) \cdot 0 \\
\end{split}
\end{equation}
$\square$