you're pretty close

$\ln\left(\dfrac{-1}{\ln(1-p)}\dfrac{p^y}{y}\right)=\ln\left(\dfrac{-1}{\ln(1-p)}\right)+y\ln(p)-ln(y)$

by inspection

$\phi=1$

$\theta=\ln(p)$

$b(\theta)=-\ln\left(\dfrac{-1}{\ln(1-p)}\right)=-\ln\left(\dfrac{-1}{\ln(1-e^\theta)}\right)$

$c(y,\phi)=-\ln(y)$