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)
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.