Enqueue is done at the rear.
Dequeue is done at the front.
So we need to point to a node such that we can access both the front and the rear easily (in $O(1)$ time).
In circular LL, if we point to the rear, we can access rear and front easily. So, Option D
// front == (rear + 1) mod n.
Why are other options incorrect?
If we point to the front, to access the rear node, we have to traverse all the way to the other end, ie, $O(n)$
So, Option B
By extension Option A
And obviously since rear node is the right answer, Option C