If C1 is the only one that has W, we start with
C1: W __ +
C2: __ __ +
C3: __ __ +
Did W have a rule?
Yes, it was that W must appear with N. Okay, so we know N must also be in C1.
C1: W N +
C2: __ __ +
C3: __ __ +
What else do we know?
We know that everything in C2 is also in C1 and that we must have 2-3 items in each C. That means that we need to add at least 2 items to C2 without putting more than 3 items in C1, which already contains W and N.
Could we simply add WN to C2? Nope. The problem stated that C1 is the only one with W. But we can add just N, since the rule is that W must have an N with it, not the other way around.
So now we have this:
C1: W N +
C2: N __ +
C3: __ __ +
What other rules is there? Only one C does P, and P can not be with G. Well we can't add P to C2 because that would copy it to C1 and it can only appear once. We can't add P to C1 because that would put it at 3 items and we still need to add another item to C2 that can be copied to C1. So that leaves C3 for P.
So now we have this:
C1: W N +
C2: N __ +
C3: P __ +
Since G can't go in C3, and whatever goes in C2 must also go in C1, then we can add G to both C1 and C2.
C1: W N G
C2: N G +
C3: P __ +
Now what materials are missing? T. We can add T to C3 to ensure that it reaches the 2 item minimum.
C1: W N G
C2: N G
C3: P T
That means that (A) is the correct answer.