Isothermal process gives V = A(T)exp(-KP)?
$A$ is already an implicit function of $T$, so your answer is essentially correct.

Starting from the definition of $\kappa$ and integrating both sides with respect to $P$, we obtain $$-\int\kappa\,\mathrm{d}P = \int\frac{1}{V}\frac{\partial V}{\partial P}\,\mathrm{d}P,$$ which yields as solution $-\kappa P = \ln V + f(T)$, where $f(T)$ is our "constant" of integration. Rearranging and exponentiating yields $V(P,T) = e^{-f(T)}e^{-\kappa P} = A(T)e^{-\kappa P}$, $A(T) := e^{-f(T)}$, and we are done.