Godly ways to get back !

First of all we check out the number of ways from $A$ to $D$ .

From $A$ to $C$ via $B$ there are : $3 \times 3 = 9$ ways

Hence, from $A$ to $C$ there are: $9 + 1 = 10$ ways

Now, from $A$ to $D$ via $C$ and $B$ we have: $10 \times 2 = 20$ ways

Hence, from $A$ to $D$ there are: $20 + 1 = 21$ ways

Now, Similarly we'll have $21$ ways back to $A$ .

Thus, there will be $21 \times 21 = 441$ ways from $A$ to $D$ and back to $A$ .

ANSWER : $\boxed{441}$

$For\quad A\quad ->\quad B\quad ->\quad C\quad ->\quad D:\\ \qquad there\quad are\quad 3\quad \times \quad 3\quad \times \quad 2\quad =\quad 18\quad ways\\ For\quad A\quad ->\quad C\quad ->\quad D:\\ \qquad there\quad are\quad 1\quad \times \quad 2\quad =\quad 2\quad ways\\ For\quad A\quad ->\quad D:\\ \qquad there\quad are\quad 1\quad ways\quad =\quad 1\quad way\\ For\quad total\quad there\quad are:\\ \qquad 18\quad +\quad 2\quad +\quad 1\quad =\quad 21\quad ways\quad from\quad A\quad ->\quad D\\ \\ For\quad back\quad D\quad ->\quad A\quad there\quad are\quad the\quad same\quad 21\quad ways\\ \\ So\quad from\quad A\quad ->\quad D\quad and\quad then\quad D\quad ->\quad A:\\ \qquad there\quad are\quad 21\quad \times \quad 21\quad =\quad \left\lceil 441 \right\rceil \quad ways$