1) Both $P$ and $Q$ equal to zero will do. A little less obvious example is when $Q$ is a function of $y$ and $P$ is a function of $x$. Anyway, the most general case is when $(P, Q) = \nabla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y})$ where $F$ is some nice enough function.
2) It is possible to find $(P, Q)$ with $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \ne 0$ such that the integral over some domain is $0$. However, if you consider any regions, you can squeeze them as small as possible around the point where $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is not zero, and that integral will be non-zero (assuming $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is continuous).
One can use Green's Theorem if the offending point, say $\vec r_0$, is excluded from the region of integration.
To do that, we deform the boundary contour with a "keyhole" contour that encircles the excluded point $\vec r_0$.
This reduced the problem to evaluating the line integral
$$\lim_{\epsilon \to 0}\int_0^{2\pi}\vec F(\vec r_0+\epsilon \hat r)\cdot \hat \phi \epsilon \, d\phi \tag 1$$
For example, suppose $\vec F(\vec r)=\frac{-\hat xy+\hat yx}{x^2+y^2}$.
NOTE:
The vector $\vec F$ here represents the static magnetic field $\vec H$ (modulo the constant $I/2\pi$) from a line current $I$ aligned along the $z$-axis.
Clearly, for all $(x,y)\ne (0,0)$, the first partial derivatives of $\vec F$ satisfy the relationship
$$\frac{\partial F_y}{\partial x}=\frac{\partial F_x}{\partial y} \tag 2$$
NOTE:
We remark that $(2)$ is the $z$ component of the curl of $\vec F$, which by Maxwell's Equation for static magnetic fields is $\hat z\cdot \nabla \times \vec H=0$ for $\vec r \ne 0$.
However, $\vec F$ is singular at the origin. We can still use $(2)$ in Green's Theorem to evaluate the line integral of $\vec F$ on any contour $C$ that encloses the origin by using the contour deformation given in $(1)$. Proceeding we find with $\vec r_0=0$
$$\begin{align}
\oint_C \vec F(\vec r)\cdot \,d\vec \ell&=\lim_{\epsilon \to 0}\int_0^{2\pi}\vec F(\vec r_0+\epsilon \hat r)\cdot \hat \phi \epsilon \, d\phi\\\\
&=\lim_{\epsilon \to 0}\int_0^{2\pi} \left(\frac{\hat \phi}{\epsilon}\right)\cdot \hat \phi \epsilon \,d\phi\\\\
&=2\pi
\end{align}$$
as expected from Ampere's Law (after multiplying by $I/2\pi$).
Best Answer
You’re making a common mistake: when the domain isn’t simply connected, being irrotational doesn’t always mean that the vector field is conservative. In this case, there’s a hole in the domain at the origin, so the integral along a closed path that surrounds this hole might not vanish.