Certainly if P is false, (P&Q) cannot be true.
But how to prove this using natural deduction?
I'd propose as a direct proof the following derivation :
(1) ~P ( Premise )
(2) ~P v ~Q ( v - intro)
(3) ~ ( P & Q) ( DeMorgan)
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1) $\lnot P$ --- premise
2) $(P \land Q)$ --- assumed [a]
3) $P$ --- from 2) by $(\land \text E)$
4) $\bot$ --- from 1) and 2), by $(\lnot \text E)$ (alternatively, using $(\to \text E)$, if $\lnot P$ is defined as $P \to \bot$)
5) $\lnot (P \land Q)$ --- from 2) and 4) by $(\lnot \text I)$, discharging [a].