Saudi Arabian Mathematical Competitions, 2015

Let $\mathcal{L}$ be the set of all the languages used in KAUST.

a.
Assume $2k \geq n+1$. Consider $P \in A_{k}$, and let $\mathcal{L}_P \subseteq \mathcal{L}$ be the set of all languages spoken by $P$. The set $\mathcal{L}_P$ contains at least $k$ languages.

Let $\mathcal{L}' \subseteq \mathcal{L}$ be a set of $k$ languages. We have
$$|\mathcal{L}_P \cap \mathcal{L}'| = |\mathcal{L}_P| + |\mathcal{L}'| - |\mathcal{L}_P \cup \mathcal{L}'| \geq 2k - |\mathcal{L}| = 2k - n \geq 1.$$
This means that $\mathcal{L}_P \cap \mathcal{L}' \neq \emptyset$ and therefore $P$ can speak at least one language in $\mathcal{L}'$. Hence $P \in B_{k}$.

b.
Assume $2k \leq n+1$. Consider $P \in B_{k}$, and let $\mathcal{L}_P \subseteq \mathcal{L}$ be the set of all languages spoken by $P$ and $\mathcal{L}' = \mathcal{L} \setminus \mathcal{L}_P$. Because there is no language in $\mathcal{L}'$ spoken by $P$ we have $|\mathcal{L}'| \leq k-1$. We deduce that
$$|\mathcal{L}_P| = |\mathcal{L}| - |\mathcal{L}'| \geq n - (k-1) \geq k.$$
This means that $P$ speaks at least $k$ languages and therefore $P \in A_{k}$.