There are basically two important things to think about here: first, the fact that y could be either even or odd; and second, the fact that multiply any two even numbers together will always yield a multiple of 4 (because those even numbers will be expressible as 2m and 2n for some integers m and n, and their product 2m*2n will then simplify to 4mn).
So:
(A) 2y(y+1)(y-1)
if y is even --> 2y will already be a multiple of 4, so we won't care about the rest.
if y is odd --> y+1 and y-1 will both be even, so when we multiply them together we'll get a multiple of 4 and not care about the rest.
(B) y(2y+2)(y-3)
if y is even --> y will be even as will 2y + 2; even times even is a multiple of 4.
if y is odd --> 2y + 2 will still be even (because 2y is even for any integer y), as will y-3, so again, we'll get a multiple of 4.
(C) y(y+3)(2y - 4)
if y is even --> y and 2y-4 will be even; their product will be a multiple of 4.
if y is odd --> y + 3 will be even and 2y-4 is even regardless of y's parity; even times even yields a multiple of 4.
(D) 2y(y+4)(y - 2)
if y is even: 2y and y+4 will be even, so we'll already have a multiple of 4.
if y is odd: TROUBLE! 2y will still be even, but y+4 and y-2 will both be odd; we'll therefore wind up with [(even)*(odd)]*(odd) = even * odd = even, but not necessarily a multiple of 4. To prove it with an example, try y = 3. Then we'll have 6(7)(1)=42, which is not divisible by 4.
We can stop there.
PS
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