To convert to conjugtive normal form we use the following rules:
Double Negation:
- $P\leftrightarrow \lnot(\lnot P)$
De Morgans Laws
$\lnot(P\bigvee Q)\leftrightarrow (\lnot P) \bigwedge (\lnot Q)$
$\lnot(P\bigwedge Q)\leftrightarrow (\lnot P) \bigvee (\lnot Q)$
Distributive Laws
$(P \bigvee (Q\bigwedge R))\leftrightarrow (P \bigvee Q) \bigwedge (P\bigvee R)$
$(P \bigwedge (Q\bigvee R))\leftrightarrow (P \bigwedge Q) \bigvee (P\bigwedge R)$
So lets expand the following
$((P\bigwedge \lnot N)\bigvee (N∧\lnot P))\bigvee Z$
$((((P\bigwedge \lnot N)\bigvee N)\bigwedge ((P\bigwedge \lnot N)\bigvee \lnot P)))\bigvee Z)$
$(((P\bigvee N)\bigwedge (\lnot N\bigvee N ) \bigwedge (P\bigvee \lnot P)\bigwedge (\lnot N \bigvee \lnot P)) \bigvee Z)$
Then noting that $(\lnot N\bigvee N)$ and $(P\bigvee \lnot P)$ are always true we may remove them and get (It's cancelling these terms that gives a 2 term answer):
$((P\bigvee N)\bigwedge (\lnot N \bigvee \lnot P)) \bigvee Z)$
$((P\bigvee N\bigvee Z)\bigwedge (\lnot N \bigvee \lnot P \bigvee Z))$
This will in general not happen though and you may get more terms in your formula in CNF. Just so you know you can also check these things on Wolfram Alpha