$$(((p\iff q)\iff r)\wedge(q\iff r))\implies p$$
Should be done with natural deduction, but up to $60$ steps. If I assume $\neg\:p$ probably I may finish up to $20$ steps, but without assumption I didn't find good way. If start could be separated by deduction rules and in the end also collect by same rules.
$$ \begin{aligned} &1.& &\neg\: ((((p\iff q)\iff r)\land (q\iff r))\implies p)&\hspace{20ex} & \\ &2.& &(((p\iff q)\iff r) \land (q\iff r))& & \\ &3.& &\neg\: p& -&\text{ assume } \\ &4.& &(p\iff q\iff r) \land (q\iff r)& -&\text{ from } 4 \\ &5.& &(p\iff q\iff r) \land E,4 & & \\ &6.& &(q\iff r) \land E,4 & & \\ &7.& &p\implies (q\iff r) & -&\text{ from } 4 \\ &8.& &(q\iff r) \implies p & -&\text{ from } 4 \\ &9.& &p(\implies E 6,8) & & \end{aligned} $$
where $\land$ is negation