I have to find a new theory by skolemization and prove that it is incosistent by tableau method.
My result:
My question is, if I have one true and one false sign with same relation and the elements are not in the same place, does it lead to contradictory? (right branch of the tree in the picture)
No, $p(c,f(c))$ is not the same statement as $P(f(c),c)$, so the two statements you circled in red do not lead to a contradiction.
I also note a mistake you make earlier in the tree: a statement like $\forall x (P(x) \lor Q(x))$ does not branch into $\forall x \ P(x)$ and $\forall x \ Q(x)$, but this is basically what you did when decomposing $\forall x \forall z (\neg R(x) \lor P(f(x),z))$. Instead, you should first instantiate the quantifiers with something, and then branch once you have isolated the $\lor$.
