Just the simple logic is that we have 2nCn/(n+1) unlabelled trees , now each unlabelled tree will always correspond to one BST , consider 1,2,3 now say you draw a right skewed unlabelled tree , now u can label it only in one way 1 2 3 to form a BST , if say u draw a left skewed unlabelled tree , then u can label it as 3 2 1 , to form BST therefore each unlabelled tree gives rise to only 1 BST.
But in the question it is already given that we are provided with one unlabelled tree therefore there is only 1 way in which we can populate it to form a BST therefore answer is option A .