Q:

How do I simplify(expand) this Boolean expression?

Expression - (A OR B OR C OR D) AND (!B AND !D)
I know that with distributive property, it holds that (a OR b) AND (c OR d) = (a AND c) OR (a AND d) OR (b AND c) or (b AND d) but I'm not sure how it will work if the second group has an AND
Steps in the answer would help.

A:

OK, so you have:
$(A \lor B \lor C \lor D) \land (\neg B \land \neg D)$
Now, notice that this is a conjunction (an 'AND' statement) whose right conjunct is a conjunction itself. In a case like that, you can actually drop the parentheses, and thus get:
$(A \lor B \lor C \lor D) \land \neg B \land \neg D$
This is what is called a generalized conjunction; a conjunction that can conjunct together any number of terms. In this case, you have $3$ terms
And by the way, note that the left conjunct is already a generalized disjunction (of $4$ terms), so this is really nothing unusual.
Now, any logical equivalence principle you would normally apply to a 'normal' conjunction with $2$ conjuncts, you can still apply to any $2$ conjuncts of a generalized conjunction.
For example, one very useful simplification principle is:
Reduction
$\neg P \land (P \lor Q) \Leftrightarrow \neg P \land Q$
To understand this: note that $\neg P$ is definitely true ... so how can $P \lor Q$ be true as well? Clearly it must be that $Q$ is true. So, the left side implies the right side, and going from right to left is trivial, since $Q$ implies $P \lor Q$.  But, going from left to right, which is how we typically use Reduction, because is simplifies things, we say that in the presence of $\neg P$, the term $P \lor Q$ 'reduces' to just $Q$
OK, so we can apply Reduction to your statement twice:  The $\neg B$ will reduce the $A \lor B \lor C \lor D$ term to $A \lor C \lor D$, and the $\neg D$ term will further reduce that to $A \lor C$. Hence, we end up with:
$(A\lor C) \land \neg B \land \neg D$
If you want this more formally, you can state the Association principle, which states that $P \land (Q \land R) \Leftrightarrow (P \land Q) \land R$, and thus makes clear that you could write $P \land Q \land R$ without any ambiguity, as a justification for dropping the parentheses at the start. As such, it formally looks like:
$$(A \lor B \lor C \lor D) \land (\neg B \land \neg D)$$
$$\overset{\text{Association}}{\Leftrightarrow}$$
$$(A \lor B \lor C \lor D) \land \neg B \land \neg D$$
$$\overset{\text{Reduction x 2}}{\Leftrightarrow}$$
$$(A\lor C) \land \neg B \land \neg D$$

