It is stated in Jacod/Shyraev: Limit theorems of stochastic processes that the predictable projection of a jump process $\Delta A$ of a process $A\in\mathcal{A}^+$ (see below for the notation) is equal to the jump process of the compensator of A, that is it is equal to $\Delta (A^p)$.
In order to show this, there is a hint to use that for a local martingale $X$, the predictable projection of X given by ${}^p X$ is equal to $X_{-}$ and ${}^p(\Delta X)=0$.
So we have $$\Delta (A-A^p)=A-A^p-A_{-}+(A^p)_{-}=\Delta A-\Delta A^p$$ which yields $$0={}^p(\Delta A)-{}^p(\Delta A^p)$$, thus $${}^p(\Delta A)={}^p(\Delta A^p)$$.
It remains to show that ${}^p(\Delta A^p)=(\Delta A^p)$, but i am not sure how to do this?
Notation: Let $\nu^+$ be the set of all real valued processes $A$ that are càdlàg, adapted with $A_0=0$ and whose each path is non-decreasing.
Let $\mathcal{A}^+$ the set of all processes $A\in\nu^+$ which satisfy $$E[A_\infty]<\infty$$ where $\text{Var}(.)$ is the variation of a stochastic process.
${}^p(\Delta A^p)=\Delta A^p$, since $\Delta A^p=A^p-A^p_-$ is predictable.