An impressive combinatorial identity I'm trying to show that, $\forall n\in\mathbb{N},\forall p\in\mathbb{N^*},$ $$\sum_{k=1}^{n+1}\left(\frac{4^k}{2k-1}{2k\choose k}\frac{\displaystyle{n+k-1\choose n-k+1}{2(n+p-k)\choose n+p-k}}{\displaystyle{n+p\choose k}}\right)=\frac{8}{p}{2(p-1)\choose p-1}\frac{\displaystyle{2(2n+p)\choose2n+p}}{\displaystyle{2n+2p\choose2n+p}}$$
I have checked numerically, I'm confident it's true. But I have absolutely no idea as of how to even start to prove this.
Any suggestion ?

Let me add a picture of maple output as well:

In the output, $N-1$ is the annihilating operator of the sum where $N$ is the shift operator with respect to $n$. The other part of the output is not important for the sake of this question.
