I think it is always a singular point in $Hom(\Gamma, SU(p,q))$. Suppose $G$ is reductive and $\frak{g}$ its Lie algebra. Let $\rho \in Hom(\Gamma, G)$. Then $\rho$ induces a $\Gamma$-module structure on $\frak{g}$ by the Adjoint action of $G$ on $\frak{g}$. Then the Zariski tangent space at $\rho$ is the group-cocycle space $T := Z^1(\Gamma, \frak{g})$. The dimension of $T$ is $$(2g - 1)dim(G) + dim(Z(\rho)),$$ where $g$ is the genus of the surface associated with $\Gamma$ and $Z(\rho)$ is the centralizer of the image of $\rho$.

In your case, $dim(Z(\rho)) = 1$ while at a smooth point, $dim(Z(\rho)) = 0$. So $\rho$ is singular.

I think it is always a singular point in $\mathrm{Hom}(\Gamma, \mathrm{SU}(p,q))$. Suppose $G$ is reductive and $\frak{g}$ its Lie algebra. Let $\rho \in \mathrm{Hom}(\Gamma, G)$. Then $\rho$ induces a $\Gamma$-module structure on $\frak{g}$ by the Adjoint action of $G$ on $\frak{g}$. Then the Zariski tangent space at $\rho$ is the group-cocycle space $T := Z^1(\Gamma, \frak{g})$. The dimension of $T$ is $$(2g - 1)\dim(G) + \dim(Z(\rho)),$$ where $g$ is the genus of the surface associated with $\Gamma$ and $Z(\rho)$ is the centralizer of the image of $\rho$.

In your case, $\dim(Z(\rho)) = 1$ while at a smooth point, $\dim(Z(\rho)) = 0$. So $\rho$ is singular.