**Option B is the most correct answer**. (To know why it is not THE correct answer read the tail section of this answer)

The reasons are as follows.

A problem *p* in NP is NP-complete if every other problem in NP can be transformed into *p* in polynomial time. They have another property that given a problem in NP and a solution, we can verify deterministically in polynomial time that the solution is indeed a valid solution or not.

P represents the class of all problems whose solution can be found in polynomial time.

Reduction of an NPC problem to a P problem would imply 2 things:

1) All problems in NP are reducible to this problem in P because all NP problems are reducible to NPC problems.

2) We can solve the NPC problem in polynomial time as well because we ca solve ther problem in P in polynomial time by definition of class P.

**This implies that all problems in NP are solvable in polynomial time**. (1)

A problem which can be solved in polynoimial time is essential in P

Which implies **P=NP** (QED)

Tail

The statement of option B is incomplete. It is not enought to say that that reduction of NPC problem to a problem in P, but the reduction should be polynomial reduction or karp reduction.

The statement which implies P=NP would be **There exist a polynomial time reduction of a problem in NPC to a problem in P.** The answer is partially correct because when we say reduction, we usually mean polynomial time reduction, but neverthless it doesnt hurt to be precise.