Proving the equivalence without making use of Truth Tables

The equivalence you have written does not hold.

Counterexample: $A$ and $B$ are both True. Then the left hand side evaluates to $(F \lor T) \lor (F \lor F)=T \lor F=T$, while the right hand side is $F$.

What does hold is that $(\neg A \lor B) \color{red}\land (\neg B \lor \neg A) \equiv \neg A$

Assuming you made a typo, and that that is what you needed to show, it is easily shown as follows:

$$(\neg A \lor B) \land (\neg B \lor \neg A) \overset{Commutation}{\equiv}$$

$$(\neg A \lor B ) \land (\neg A \lor \neg B)\overset{Distribution}{\equiv}$$

$$\neg A \lor (B \land \neg B) \overset{Complement}{ \equiv}$$

$$\neg A \lor \bot \overset{Identity}{\equiv}$$

$$\neg A$$