The semaphore $\verb|S|$ is initialized to $3$ , so at most $3 ~\verb|wait(S)|$ operations can succeed simultaneously. Hence, at most $3$ processes can be in the critical section at any time.
Option A is correct.
Deadlock is not possible because every process that enters the critical section eventually executes $\verb|signal(S)|$, and there is no circular wait.
Option B is false.
Starvation is possible if the semaphore implementation uses unfair scheduling. A process may remain waiting indefinitely while others repeatedly enter the critical section.
Option C is correct.
It is not guaranteed that exactly 3 processes will always be in the critical section, since fewer than $3$ may be executing or ready at some times.
Option D is false.
Correct options: A, C