How to prove this using natural deduction $$⊢ P ∨ ¬P$$ 
I found this question on the net. I know the solution, but I find it complicated. 
How should I approach this sort of question? Or can you provide me with another solution?


I don't know, whether you really find this helpful, but you could prove a bunch of other (generally useful) statements / rules first: 


*

*$\neg(p \wedge q) \Rightarrow \neg p \vee \neg q$ (DM)

*$\neg(p \wedge \neg p)$ (PNC)

*$(p \vee q, p\vdash p', q \vdash q') \vdash p' \vee q'$ (CDL)


and then it should be easy:


*

*$\neg(p\wedge \neg p)$ (by PNC)

*$\neg p \vee \neg \neg p$ (by DM, 1.)

*$\neg p \vdash \neg p$ (Assumption rule)

*$\neg \neg p \vdash p$ (by $\neg\neg E$)

*$p \vee \neg p$ (by CDL, 2.,3.,4.)
Here, for example, the proof for (PNC):


*

*$p \wedge \neg p$ (H)

*$\,\,\,\,$ $p$ $\,\,\,\,$ ($\wedge E1, 1.$)

*$\,\,\,\,$ $\neg p$ $\,\,\,$ ($\wedge E2, 1.$)

*$\,\,\,\,$ $\perp$ $\,\,\,\,$ ($\Rightarrow E$, 2., 3.)

*$\neg(p \wedge \neg p)$ ($\Rightarrow I$, 1., 4.)

