Since you know that $P(i)=e_iA$, try to understand when this is nonzero. Let $p \in A$ be a path, then $e_i p \neq 0$ iff $p$ starts in $i$. This way you get a basis of this space. So, $P(1)=\langle e_1\rangle_K$, $P(2)=\langle e_2, \beta,\delta\rangle_K$, and $P(3)=\langle e_3, \alpha, \gamma, \alpha\delta, \gamma\beta\rangle_K$ (since $\alpha\beta$ and $\gamma\delta$ are relations).
The action of the arrows is just by concatenation (as usual in the path algebra).
The next step is to remember the equivalence between modules and representations: Whenever you have a module $M$, it is mapped to the representation given by the vector space $Me_j$ on each arrow, and the linear map $-\cdot a: Me_j\to Me_k$ for an arrow $a:j\to k$.
In the example of $P(2)$ you get
and in the example of $P(3)$ you get:
(From this step, it is also easy to "see" the corresponding radicals.
As a final step, translate this to $K^r$ and matrices by using an explicit isomorphism to the claimed representations.