If k is a positive integer, then is even only when k is even even

The product contains k and k+1, which are consecutive integers. One of those two consecutive integers must be even, so we have an even number in our product, and the entire product will be even no matter what. So the answer can only be D or E.

Looking at E, if k is odd, then k+1 and k+3 are both even, so both are divisible by 2, and their product will be divisible by 4. So E is correct. In fact, since k+1 and k+3 are consecutive even integers, one of them must be divisible by 4, so the product will be divisible by 8, but we don't need that here.

If you wanted to know when this product is divisible by 3 (to see why D is wrong), you can notice that k, k+1 and k+2 are three consecutive integers. Among any three consecutive integers, we always find exactly one multiple of 3. Since k and k+1 are in our product, our product will be divisible by 3 any time either of those numbers is divisible by 3. But k+2 is not in our product, and if k+2 is divisible by 3, then k, k+1 and k+3 will not be. So our product will fail to be a multiple of 3 precisely when k+2 is divisible by 3, so the product (k)(k+1)(k+3) will not be divisible by 3 if k = 1, 4, 7, 10, 13, etc. It's irrelevant whether k is even or odd, so D is wrong.