It might help to think by analogy with algebraic expressions formed using multiplication and addition. Multiplication distributes through addition, i.e., you have $x \times (y + z) = (x \times y) + (x \times z)$. Hence you can write any expression as a sum of products of atoms, by pushing the multiplications in through additions. But note that addition does not distribute through multiplication, so you can't write every expression as a product of sums of atoms.
Boolean algebra, however is much more symmetric: not only does conjunction distribute through disjunction, i.e., $x \land (y \lor z) = (x \land y) \lor (x \land z)$ but also disjunction distributes through conjunction, i.e., $ x \lor (y \land z) = (x \lor y) \land (x \lor z)$. Hence (also using De Morgan's laws to push in negations), you get to choose whether to push the conjunctions in, which will lead you to DNF: a disjunction of conjunctions of literals, or to push the disjunctions in, which will lead you to a CNF: a conjunction of disjunctions of literals. (Here "literal" means an atom or negated atom.) For example, your formula can be transformed into a CNF like this:
$$
\begin{array}{rcl}
[(p \land q) \lor \lnot p] \lor \lnot q &=& [(p \lor \lnot p) \land (q \lor \lnot p)] \lor \lnot q\\
&=& (p \lor \lnot p \lor \lnot q) \land (q \lor \lnot p \lor \lnot q)
\end{array}
$$
You can now do further simplifications if you wish, e.g., to arrive at $p \lor \lnot p$ (which is both a DNF and a CNF).
