How to convert (B ∨ ¬A) → (C ∨ ¬D) to conjunctive normal form? I have tried solving it. But, I am stuck:
(B ∨ ¬A) → (C ∨ ¬D) // removing implication
¬(B ∨ ¬A) V (C ∨ ¬D) // using De Morgan's Laws
(¬B ∧ A) V (C ∨ ¬D) // This is not CNF. What to do next?

 A: Use Distribution, which in general says that $P \lor (Q \land R) \Leftrightarrow (P \lor Q) \land (P \lor R)$
Applied to your last statement:
$(\neg B \land A) \lor (C \lor \neg D) \Leftrightarrow$
$(\neg B \lor C \lor \neg D) \land (A \lor C \lor \neg D)$
A: $(B \lor \lnot A) → (C \lor \lnot D)\label{1}\tag{1}$ 
$(\lnot B \land A) \lor C \lor \lnot D\label{2}\tag{2}$ 
The best way to answer these types of questions is by Karnaugh Map. If you are not familiar with them there is a page here that explains them quite well 
https://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Logic/Logic3.html
Take your simplified expression ($\ref{2}$) and draw it out on a Karnaugh Map. Next; select regions of the map in such a way as to OR statements together. Next choose one or more statements to AND your first statement together with to produce the final expression.
I choose to start by writing ($\ref{3}$).
$A\lor C\lor \lnot D\label{3}\tag{3}$
Then I realized that there was one undesirable region on the map that ($\ref{3}$) picked up (namely ($\ref{4}$) ).
$A\land D\land B\land\lnot C\label{4}\tag{4}$
To remove this case, I AND'ed my previous expression ($\ref{3}$) with the negation of ($\ref{4}$). The result was
$(A\lor C\lor \lnot D)\land(\lnot A\lor \lnot B\lor C\lor \lnot D)\label{5}\tag{5}$
Which is equivalent to your original expression ($\ref{1}$) except in CNF (Conjunctive Normal Form)
