I believe you meant to type "curve of degree m". Let's prove the "only if" direction first. If you can prove it for triangles then you can triangulate your $n$-gon and multiply together the expressions for each triangle. So let's assume $n=3$.
Now, pick coordinates $(t_0,t_1,t_2)$ on $\mathbb P^2$ to correspond to the three lines of your triangle, and let $at_0+bt_1+bt_2$ be a generic line that doesn't pass through any of your points. Consider the three functions $$f_i=\frac{t_i}{at_0+bt_1+ct_2}$$ on your degree $d$ curve. Weil reciprocity then tells you that $\prod_{P} (f_i,f_j)_P=1$ so you can write $$\prod_P (f_0,f_1)_P(f_1,f_2)_P(f_2,f_0)_P=1$$ and by the definition of the Weil symbol this ends up being exactly the product $\prod_{i=1}^n \prod_{j=1}^m \frac{\overline{B_{ij}A_i}}{\overline{B_{ij}A_{i+1}}}$ in your statement. I learned this proof here.
To got the other way, omit one of the points and pick a degree $d$ curve through the remaining $3d-1$, by the calculation above the remaining point will be uniquely determined.