There can be two answers depending on how complex & deep we want to go.
CASE 1: Considering in the question they are implementing mutex exactly same as a binary semaphore, then:
Yes, after P10 performs signal(or V) on 'mutex' it makes/keeps the value of mutex to/as 1 and exits. After that, if any other process in the critical section perform signal on 'mutex' then that process will also exit by keeping the value of mutex as 1.
P1 to P9 has: P10 has:
P(mutex) /* same as wait(mutex) */ V(mutex)
V(mutex) /* same as signal(mutex) */ V(mutex)
Considering binary semaphore here. If we trace out then we will see that at max we can have all 10 processes in the CS at a moment because we can use P10 repeatedly. In one iteration of P10 it is allowing two processes from P1-P9 to get inside the CS.
CASE 2: The thing is binary semaphore is different than mutex. See this: https://en.wikipedia.org/wiki/Semaphore_(programming)#Semaphores_vs._mutexes
This means P() is saving process-id of the process performing the operation. P10 will never enter CS because it is not locking mutex using P() such that its process-id is being saved by the mutex. Since P10 is performing V() first, it will be denied as process-ids won't match, indicating some problem in the implementation. And thus, we will have only one process inside the CS at any moment.
Generally, Case 1 will be your answer for any written exams. But in interviews, Case 2(along with Case 1) will look better.