Answer is (**C**) .

Reasoning :-

First let me explain what is counting semaphore & How it works. Counting semaphore gives count, i.e. no of processes that can be in Critical section at same time. Here value of $S$ denotes that count. So suppose $S = 3$, we need to be able to have $3$ processes in Critical section at max. Also when counting semaphore $S$ has negative value we need to have Absolute value of $S$ as no of processes waiting for critical section.

(A) & (B) are out of option, because $Xb$ must be $1$, otherwise our counting semaphore will get blocked without doing anything. Now consider options (C) & (D).

Option (D) :-

$Yb = 1, Xb = 1$

Assume that initial value of $S = 2$. (At max $2$ processes must be in Critical Section.)

We have $4$ processes, $P1, P2, P3 \& P4.$

$P1$ enters critical section , It calls $P(s) , S = S - 1 = 1.$ As $S > 1$, we do not call $Pb(Yb)$.

$P2$ enters critical section , It calls $P(s) , S = S - 1 = 0.$ As $S >0$ we do not call $Pb(Yb).$

Now $P3$ comes, it should be blocked but when it calls $P(s) , S = S - 1 = 0-1 = -1$ As $S < 0$ ,Now we do call $Pb(Yb)$. Still $P3$ enters into critical section & We do not get blocked as $Yb$'s Initial value was $1$.

This violates property of counting semaphore. $S$ is now $-1$, & No process is waiting. Also we are allowing $1$ more process than what counting semaphore permits.

If $Yb$ would have been $0, P3$ would have been blocked here & So Answer is (**C**).

$Pb(yb);$