The $p_i$s are self-loops, and special ones, since they're required to compose trivially. That is, if you look at the quiver as a category, they're the identity morphisms. Then in any representation they're mapped to the identity automorphism of each vector space.
As to your question about their sum: take an arbitrary element $\pi=\sum c_j \pi_j,$ where the $\pi_j$ are arbitrary paths in $Q$ and $c_j$ constants from whatever you're taking $P_Q$ over.
Now in considering the product $(\sum p_i) \pi,$ the claim is that each $c_j\pi_j$ is annihilated by all but one of the $p_i,$ and sent to itself by the last. This is because the paths can't concatenate if the end point of $\pi_j$ isn't the start (and end) point of $p_i$, in which case the product is defined as 0, while the $p_i$ that does match the end point of $\pi_j$ is required to composed trivially with it. So each component path in $\pi$ comes back out of the product once, and only once.