First, disregard the constraint $p_y=q_x$. Consider the integral curves of the vector field $V:=(1,-a)$, namely the solutions of the ODE $$\frac{dy}{dx}=-a(x,y).$$ They are parametrized by $x$. Your equations are ODEs along these curves, $$\dot p=bq,\qquad\dot q=-bp.$$ Set $q+ip=\rho e^{i\theta}$. Then the ODEs become $\dot\rho=0$, $\dot\theta=b$. You see that you need an initial data over a curve transversal to $V$, for instance along a curve $x={\rm cst}$.
If we now take in account the constraint, the system becomes over-determined. Such systems usually do not have a solution, unless the data (here $a,b$ and the initial data) are very special. To see how it begins, let us set $h:=p-aq$. Then we have $h_x=(b-a_x)q$ and $h_y=-bp-a_yq$. By Schwarz, we obtain a new equation $$(a_xq-bq)_y=(a_yq+bp)_x.$$ This is a fourth differential equation. With the three others, you may solve $$(p_x,p_y,q_x,q_y)=F(p,q,x,y),$$ unless a $4\times4$ determinant vanishes. And the tale does not end there. You can continue and find other first-order PDEs, which yield incompatible in just one more step. Again, unless you are very lucky.
edit after a few hours. Once you have the first derivatives in terms of $p$ and $q$, you may apply Schwarz, $(p_x)_y=(p_y)_x$ and $(q_x)_y=(q_y)_x$. This gives $\partial_yF_1(p,q,x,y)=\partial_xF_2(p,q,x,y)$ and $\partial_yF_3(p,q,x,y)=\partial_xF_4(p,q,x,y)$. Eliminating first derivatives, there remain two equations $p=P(x,y)$ and $q=Q(x,y)$. Remark that you may not any more impose initial data, because you obtain explicit form of $p,q$ without really solving a differential equation. You have only use elimination and Schwarz identity. Finally, it happens in general that $P$ and $Q$ do not solve at all the differential equation.
In theory, you could explicit a necessary and sufficient condition in terms of $a$ and $b$ in order that your overdetermined system admit a solution. A very classical and simple situation is the system $u_x=f$, $u_y=g$, where the condition for having a solution is $f_y=g_x$.