Answer is only $\boxed{4}$.

We have, $\dfrac{p}{a}+\dfrac{p}{b}=1 \implies p=\dfrac{ab}{a+b} \implies \boxed{p(a+b)=ab}$ Now as $p | a+b$ thus, $p^{2} | ab$. Thus, $$a+b \equiv 0(\text{mod} p)$$$$\implies a \equiv -b(\text{mod} p)$$$$\implies ab \equiv -b^2(\text{mod} p)$$Hence, $p | b^{2}$ or $p | b$ because $p$ is a prime number. Also $p | a$. Let $(a,b)=(pm,pn)$ where $m,n \in \mathbb{N}$. Thus we shall get, $$\dfrac{1}{m}+\dfrac{1}{n}=1 \implies (m-1)(n-1)=1 \implies (m,n)=(2,2)$$Hence $(a,b)=(2p,2p)$ and $\dfrac{a+b}{p}=\boxed{4}$.