We have to establish whether the set P has positive multiples of 3 in it or not. Positive multiples
of 3 include {3,6,9,12...}

Statement 1
For any integer in P, (we can consider 3, as it is given in the question stem), 3+(that integer, here it is 3) is in P, so 6 is in P. Similarly 6+=9 is in P and so on. Sufficient

Statement 2
For any integer in P, 3-3=0 is in P 0-3 -3 is in P, you can see here that we have 0,-3 and so on in the set. So our set consists of {3,0,-3..} Insufficient

Great explanation, sl750!

One other thing I'd add here about problems with this phrasing - I've seen at least a few that define a set with the phrasing: if x is in the set, then ______________ is in the set.

The key to interpreting these is to realize that "x" describes "any number in the set". So in a case like this the set runs in perpetuity. For any value in the set, you can apply that definition to create a new number. So, as sl750 mentioned about Statement 1:

We know that 3 is in the set. And we know that, for any value x, x + 3 is also in the set. So 3 + 3 = 6 is in the set. And then 6 qualifies as "any number in the set" (after all, we've proven that it's a number in the set), so 6 + 3 = 9 is in the set. And now 9 qualifies as a value of x (it's "a number in the set"), so 9 + 3 = 12 is in the set, and so on.

What's often tricky is that people don't see that x means "any number in the set". But that's the definition given in the statements, so any number that we create by that definition also qualifies as a new number to which we can apply the same definition.