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?

 A: 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.)

A: $\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$
That is almost correct.   You were aiming at a proof by contradiction, and that needs to use just one subproof (also by contradiction). 
$$\fitch{}{\fitch{1.~\lnot(p\vee\lnot p)\hspace{10ex}\text{H (assumption or hypothesis)}}{\fitch{2.~p\hspace{15ex}\text{H (assumption)}}{3.~p\vee\lnot p\hspace{10ex}\text{I$\lor$ 2 (disjunction introduction)}\\4.~\lnot(p\lor\lnot p)\hspace{6.5ex}\text{IT1 (reiteration)}}\\5.~\lnot p\hspace{17.5ex}\text{I$\lnot$2,3,4 (negation introduction)}\\6.~p\vee\lnot p\hspace{14ex}\text{I$\lor$ 5 (disjunction introduction)}\\7.~\lnot(p\lor\lnot p)\hspace{10.5ex}\text{IT1 (reiteration)}}\\8.~\lnot\lnot(p\vee\lnot p)\hspace{12.5ex}\text{I$\lnot$ 1,6,7 (negation introduction)}\\9.~p\vee\lnot p\hspace{17.5ex}\text{E$\lnot\lnot$ (double negation elimination, aka DNE)}}$$
Thus the Law of Excluded Middle (LEM) is provable, iff you accept the Double Negation Elimination (DNE) rule of inference.

Note: Some rules systems combine inferences on lines 8 and 9 into one step, and call that the rule of negation elimination (or indirect proof)
$$\fitch{}{\fitch{1.~\lnot(p\vee\lnot p)\hspace{10ex}\text{H (assumption or hypothesis)}}{\fitch{2.~p\hspace{15ex}\text{H (assumption)}}{3.~p\vee\lnot p\hspace{10ex}\text{I$\lor$ 2 (disjunction introduction)}\\4.~\lnot(p\lor\lnot p)\hspace{6.5ex}\text{IT1 (reiteration)}}\\5.~\lnot p\hspace{17.5ex}\text{I$\lnot$2,3,4 (negation introduction)}\\6.~p\vee\lnot p\hspace{14ex}\text{I$\lor$ 5 (disjunction introduction)}\\7.~\lnot(p\lor\lnot p)\hspace{10.5ex}\text{IT1 (reiteration)}}\\8.~p\vee\lnot p\hspace{17.5ex}\text{E$\lnot$ 1,6,7 (negation elimination, aka indirect proof)}}$$
Thus is the Law of Excluded Middle (LEM) provable iff you allow the Indirect Proof (IP) rule of inference... or other equivalent formulations.
