(B) is correct option!
$\textbf{Wait (S)}: \text{If } S > 0 \text{ then } S \leftarrow S-1, \text{else suspend the execution of this process.}$
$\textbf{Signal (S)}: \text{If there are processes that have been suspended on semaphore} \:S,$ $\text{ then wake any one of them, else } S \leftarrow S+1$
Suppose there are only $2$ processes in the system i.e., $N=2.$
s :=1
cobegin
P(1) | P(2);
coend;
There are two concurrent processes lets say P1 and P2, which are trying to to access critical section, by executing $P(1)$ and $P(2)$ respectively.
Suppose $P(1)$ gets executed first, and it makes $s=0$. when $P(2)$ executes $Wait(S)$ then P2 will be blocked when $P1$ executes $Signal(S)$, there is one suspended process, hence $P2$ will be waken up and be placed into ready queue, now it can execute its critical section. The value of S is still 0, hence ME is satisfied.
There is only 1 process in the suspended state, hence there is no chance that it will starve because as soon as the other process performs signal operation this process will be waken up.
If $N > 2,$ then there is a chance that multiple processes are blocked, and a random process will be unblocked when a process perform $signal(S)$ operation, there is a chance that one blocked process is not getting its chance hence it might starve. So, (E) is not correct option.
PS: A blocked queue semaphore awakens processes in the order they are suspended- hence no starvation even for $N > 2$ if used in this question.