The convergence is determined by whether $q>1$. If so, then yes; if not, then no. The parameter $p$ plays no part in this determination, except (as @coco mentions below) where, when $p=0$, the integral always converges.
1
user63181
On
If $p\neq 0$
$$\frac{\arctan(px)}{x^q}\sim_\infty\frac{\pm\pi}{2x^q}$$
according to the sign of $p$ so the integral is convergent if $q>1$
and if $p=0$ the integral is convergent for all $q\in\mathbb{R}$.
The convergence is determined by whether $q>1$. If so, then yes; if not, then no. The parameter $p$ plays no part in this determination, except (as @coco mentions below) where, when $p=0$, the integral always converges.