Hint
What you have to prove is:
$(\alpha \to (\beta \to \lnot (\alpha \to \lnot \beta)))$.
In order to do this, you have to prove negation introduction: $(\phi \to \psi) \to ((\phi \to \lnot \psi) \to \lnot \phi)$.
Using it we have:
1) $\alpha, \beta, \alpha \to \lnot \beta \vdash \beta$
2) $\alpha, \beta, \alpha \to \lnot \beta \vdash \lnot \beta$ --- by MP
3) $\alpha, \beta \vdash \lnot (\alpha \to \lnot \beta)$ --- using the law above.
The result follows by Deduction Theorem.
This is not complete.
To show it is not complete, let us consider an alternative semantics for the operators involved. That is, suppose that all statements involved evaluate to either $0$, $1$, or $2$. That is, suppose all atomic variables take the value of either $0$, $1$, or $2$, suppose that $\bot$ is a constant that denotes $1$, and suppose that the $\to$ operator works as follows:
\begin{array}{cc|c}
P&Q&P\to Q\\
\hline
0&0&0\\
0&1&1\\
0&2&2\\
1&0&0\\
1&1&0\\
1&2&0\\
2&0&0\\
2&1&2\\
2&2&0\\
\end{array}
With that, we can also figure out how $\neg$ works:
\begin{array}{c|ccc}
P&P & \to & \bot\\
\hline
0&0&1&1\\
1&1&0&1\\
2&2&2&1\\
\end{array}
OK, so now let's evaluate the three axioms you have:
\begin{array}{c|ccc}
P&\neg & \neg P &\to &P\\
\hline
0&0&1&0&0\\
1&1&0&0&1\\
2&2&2&0&1\\
\end{array}
\begin{array}{cc|ccc|cc}
P&Q&\neg P & \to & (P \to Q)&P & \to & (Q \to P)\\
\hline
0&0&1&0&0&0&0&0\\
0&1&1&0&1&0&0&0\\
0&2&1&0&2&0&0&0\\
1&0&0&0&0&1&0&1\\
1&1&0&0&0&1&0&0\\
1&2&0&0&0&1&0&2\\
2&0&2&0&0&2&0&2\\
2&1&2&0&2&2&0&0\\
2&2&2&0&0&2&0&0\\
\end{array}
So notice that all of your axioms have the property that they will always evaluate to $0$, no matter what. As such, we can call them '$0$-tautologies'
Also note that if you look at the definition of the $\to$ operator, you will find that whenever $P \to Q$ has the value of $0$, and $P$ has the value of $0$, $Q$ will have to have the value of $0$ as well. This means that if you have any two $0$-tautologies, then the only kind of statement that you can infer from that using Modus Ponens is another $0$-tautology.
Finally, consider the statement $(P \to \neg P) \to \neg P$. This is not a $0$-tautology:
\begin{array}{c|ccccc}
P&(P & \to & \neg P) & \to & \neg P\\
\hline
0&0&1&1&0&1\\
1&1&0&0&0&0\\
2&2&0&2&\color{red}{2}&2\\
\end{array}
So, this means that $(P \to \neg P) \to \neg P$ cannot be inferred from your axioms and Modus Ponens. But since $(P \to \neg P) \to \neg P$ is a tautology in normal propositional logic, that means your system is not complete.
Best Answer
Hint: Remember you get to pick what $\psi$ is. Is there any formula $\psi$ such that you know $(\phi\to\psi)$ is true?
More details are hidden below.