Consider the following declaration:
int a, *b=&a, **c=&b;
The following program fragment
answer = option D
In the above figure each '*' operator corresponds to a back edge.
b is a pointer that points to a.
c is a pointer to a pointer, and the latter is b.
a = 4.
dereference c twice, we reach a. $c\rightarrow b\rightarrow a$ (One $\rightarrow$ for each $*)$
So, a = 5.