You are asked to evaluate a proof attempt of a proposition p which begins with the assumption: You are asked to evaluate a proof attempt of a proposition p which begins with the assumption: Suppose if p is false and performs a set of correct derivations and ends with the conclusion: Therefore, we conclude that p is true. Does this attempt prove the proposition p? What is(are) the propositional law(s) that you used to arrive at your decision? 
 A: If the premisses of a valid argument are true Then you can be sure that the conclusion of your argument is also true.
But,
If the Conclusion of a valid argument is False or a contradiction, Then you can be sure that your hypothesis which is (not p) in your case is False.
In other words,
$$\Bigl(\text{not} \;p \implies (q    \text{ and   not } q)\Bigr) \implies p$$
is a logically valid argument.
Some  other valid arguments are :
Modus Ponens, Modus Tollens, Disjonctive Syllogism, De Morgan's Laws,....
A: You're asking whether it's true that $ (\lnot p \Rightarrow p)$ proves  $p.$  The answer is yes, because if $\lnot p \Rightarrow p$, then because also $\lnot p \Rightarrow \lnot p$, it follows that $\lnot p \Rightarrow (p \land \lnot p)$, which proves $\lnot (\lnot p)$, which proves $p$.
A: If your proof system is based on natural deduction, then what you are given is that $\lnot P \vdash P$.  However, also by the assumption rule, $\lnot P \vdash \lnot P$.  Therefore, by the negation elimination rule $\lnot E$, we get $\lnot P \vdash \bot$.  But precisely by the rule known as $IP$ in classical natural deduction (for example in the listing on the right hand side at https://proofs.openlogicproject.org/), this implies that $\vdash P$.
In terms of a Fitch-like formulation of natural deduction, the resulting proof outline would look something like this:

