De Morgan's Law states $ \neg(a + b) \equiv \neg a\neg b $ and $\neg(ab) \equiv \neg a + \neg b$.
$$\begin{equation} \begin{aligned}A\neg B + BD \equiv & \neg(\neg(A\neg B)\neg(BD)) \text{ De Morgan's outside} \\ \equiv & \neg((\neg A + B)(\neg B + \neg D)) \text{ De Morgan's inside} \\ \equiv & \neg(\neg A \neg B + \neg A \neg D + B \neg D) \text{ Distributivity} \\ \equiv & \neg(\neg A \neg B + \neg A \neg D (\neg B + B) + B \neg D) \text{ Complementation} \\ \equiv & \neg(\neg A \neg B + \neg A \neg D \neg B + \neg A \neg D B + B \neg D) \text{ Distributivity} \\ \equiv & \neg(\neg A \neg B(1 + \neg D) + B \neg D (1 + \neg A)) \text{ Distributivity} \\ \equiv & \neg(\neg A \neg B + B \neg D) \text{ Annihilator} \\ \equiv & (A + B)(\neg B + D) \text{ De Morgan's outside}\end{aligned}\end{equation} $$
You might also want to look into K-maps.
For DNF :
a formula is a disjunctive normal form if and only if it is a disjunction of one or more conjunctions of one or more literals.
Tus, the formula :
$(\lnot z) \lor y$
is in DNF beacuse is the disjunction of two "degenerate" conjunctions : $\lnot z$ and $y$.
For CNF, we have that :
a formula is a conjunctive normal form (CNF) if and only if it is a conjunction of clauses, where a clause is a disjunction of literals.
Thus :
$((\lnot z) \lor y) \land T$
is in CNF, because we have the disjunction of the two literals : $\lnot z$ and $y$ that makes the clause : $(\lnot z) \lor y$; then a second clause : $T$ (a "degenerate" disjunction) and the two disjunctions are conjoined into : $((\lnot z) \lor y) \land T$.
Regarding the formula :
$T \land [T \lor (\lnot x \land y ) \lor ( x \land y )]$
we have three conjunctions : $(\lnot x \land y )$ and $( x \land y )$ and the "degenerate" : $T$, and they are disjoined into :
$T \lor (\lnot x \land y ) \lor ( x \land y )$;
this formula is in DNF; but adding the part : $T \land \ldots$, the resulting formula is nor more a DNF.
With the formula :
$T \land [T \lor (\lnot x \land y ) \lor ( x \land y )]$
we can apply Distributivity to its subformula:
$$[(\lnot x \land y ) \lor ( x \land y )] \equiv [y \land (x \lor \lnot x)] \equiv (y \land T) \equiv y$$
to get the equivalent :
$[T \land (T \lor y)] \equiv (T \land T) \equiv T$
which is both a CNF and a DNF equivalent to the original formula.
Best Answer
$$A \lor (B\land \lnot A) = (A \lor B) \land (A\lor \lnot A) = A \lor B$$ and $$\lnot B \lor (B \land \lnot A) = (\lnot B \lor B) \land (\lnot B \lor \lnot A) = \lnot B \lor \lnot A$$