The answer to the question in the title, "is it possible for a maximum weight edge of a cycle being included in MST?", is "not necessarily".
The correct answer to the multiple-choice question is (B).
A counterexample to (A): Let $G$ be an triangle with weight $1$, $1$ and $1$. Let $e$ be any edge of $G$.
The answer by raindrop says (A) is correct. However, he/she also points out that "A possible exception is if multiple edges have the same maximum weight, e.g. there are multiple maximum edges on $C$ which qualify to be labeled $e$. In that case, a minimum weight spanning tree may contain $e$". Let me repeat the condition about $e$ in OP's post is Let "$e$" be an edge of maximum weight on $C$. The article "an" in "an edge" does indicate that possibility. Since we are in math or computer science, one counterexample or one exception is enough to prove the fallacy of a statement.
Proof of (B): Let $m$ be an MST of $G$. If $e$ is not in $m$, we are done. Now suppose $e$ is in $m$. If we remove $e$ from $m$ (but do not remove $e$'s vertices), $m$ will be split into two trees, which we name $m_1$ and $m_2$. As a cycle that connect $m_1$ and $m_2$ by $e$, $C$ must also connect $m_1$ and $m_2$ at another edge, which we name $e'$. Then $m_1$ together with $m_2$ and $e'$ is an MST of $G$ that does not contain $e$.
A counterexample to (C): Let $G$ be an triangle with weight 1, 1 and 1. Let $e$ be any edge of $G$. The shortest path between $e$'s two vertices contains one edge, $e$ itself.
(D) is wrong since (B) is correct.