If the list is empty, $\texttt{head == NULL}$, so $\texttt{p->next}$ causes invalid access.
If the list has one node, then $\texttt{q = p->next}$ becomes $\texttt{NULL}$, so $\texttt{q->next}$ causes invalid access.
For lists with more than one node, $\texttt{p}$ reaches the second-last node and $\texttt{q}$ reaches the last node.
Then $\texttt{p->next = NULL}$ removes the last node correctly.
Answer : C