(D) is the answer.
(A) Let X = {3,6,9,8}. Let α denote multiple of 3 and β denote multiple of 4. (∀x)[α] becomes false as 8 is not a multiple of 3, and so (∀x)[α] ⇒ (∀x)[β] becomes TRUE. Now, this won't imply (∀x)[α ⇒ β] as multiple of 3 doesn't imply multiple of 4 for 3, 6 or 9.
(B) Let X = {3,6,9}. Let α denote multiple of 3 and β denote multiple of 4. Now LHS is TRUE but RHS is false as none of the x in X, is a multiple of 4.
(C) Let X = {3,6,9,7}. Let α denote multiple of 3 and β denote multiple of 4. Now (∀x)[α ∨ β] becomes false and hence LHS = ((∀x)[α ∨ β] ⇒ (∃x)[α]) becomes true. But RHS is false as 7 is not a multiple of 3.
(D) This is valid. LHS is saying that if α is holding for any x, then β also holds for that x. RHS is saying if α is holding for all x, then β also holds for all x. Clearly LHS $\implies$ RHS (but RHS does not imply LHS).
For example, let X = {4, 8, 12}, α denote multiple of 2 and β denote multiple of 4. LHS = (∀x)[α ⇒ β], is TRUE. RHS is also true. If we add '3' to X, then LHS is true, first part of RHS becomes false and thus RHS also becomes TRUE. There is no way we can make LHS true and RHS false here. But if we add 2 and 3 to X, RHS will be true and LHS will be false. So, we can't say RHS implies LHS.