Use Fitch system to proof ((p ⇒ q) ⇒ p) ⇒ p without any premise. ONLY FOR FITCH SYSTEM.

$\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$

Maybe this works for your rules?

$\fitch{}{ \fitch{ 1. (p \rightarrow q) \rightarrow p \quad Assumption}{ \fitch{ 2. \neg p \quad Assumption}{ \fitch{ 3. p \quad Assumption}{ \fitch{ 4. \neg q \quad Assumption}{ 5. p \quad Reiteration \ 3}\\ 6. \neg q \rightarrow p \quad \rightarrow Intro \ 4-5\\ \fitch{ 7. \neg q \quad Assumption}{ 8. \neg p \quad Reiteration \ 2}\\ 9. \neg q \rightarrow \neg p \quad \rightarrow Intro \ 7-8\\ 10. \neg \neg q \quad \neg Intro \ 6,9\\ 11. q \quad \neg Elim \ 10}\\ 12. \quad \quad p \rightarrow q \quad \rightarrow Intro \ 3-11\\ 13. \quad \quad p \quad \rightarrow Elim \ 1,12}\\ 14. \quad \neg p \rightarrow p \quad \rightarrow Intro \ 2-13\\ \fitch{ 15. \quad \quad \neg p \quad Assumption}{ 16. \quad \quad \neg p \quad Reiteration \ 15}\\ 17. \quad \neg p \rightarrow \neg p \quad \rightarrow Intro \ 15-16\\ 18. \quad \neg \neg p \quad \neg Intro \ 14,17\\ 19. \quad p \quad \neg Elim \ 18}\\ 20. ((p \rightarrow q) \rightarrow p) \rightarrow p \quad \rightarrow Intro \ 1-19 }$