= Solution
For a sufficiently smooth vector field on an oriented surface $S$ with compatible boundary orientation, <Stokes theorem> states
$$
\oint_{\partial S}\mathbf F\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS.
$$
Apply it to $\mathbf F=(P,Q,0)$ in the plane. The closed <line integral> is the area <integral> of $Q_x-P_y$. Thus $P_y=Q_x$ makes it vanish whenever a spanning surface lies within the smooth domain. In a simply connected domain this implies <path independence>: join one path from $A$ to $B$ to the reverse of another, and their difference is a zero closed <line integral>.
The domain qualification is essential. On the punctured plane, $P=-y/(x^2+y^2)$, $Q=x/(x^2+y^2)$ satisfy $P_y=Q_x$, but the <line integral> around the unit circle is $2\pi$. A disk spanning that circle passes through the singularity. Hence the printed implication requires an appropriate domain assumption; zero <curl> alone is not a global <path independence> theorem.
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