Step 1) Let the numbers of edges starting in all verticles be $k$. Let's think about the set below: $$X=\{(edge A, verticle B)\mid B\subset A\}$$Then, $$\mid X\mid =2q=pk\rightarrow k=\frac{2q}{p}$$Step 2) Let's call the sum of the numbers of its ends and of the edge itself $"eg"$. Define $S$ as the sum of all $"eg"$s. Then $$S=qs=k\cdot (1+2+\cdots +p)+((p+1)+(p+2)+\cdots +(p+q))=\frac{2q}{p}\cdot \frac{p(p+1)}{2}+\frac{(2p+q+1)q}{2}$$Which is equivalent to the eqation above.