15. As shown in Figure 5, the line $y=-\frac{\sqrt{3}}{3} x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. An isosceles right triangle $\triangle A B C$ with $\angle B A C=90^{\circ}$ is constructed in the first quadrant with segment $A B$ as one of its legs. If there is a point $P\left(a, \frac{1}{2}\right)$ in the second quadrant, and the area of $\triangle A B P$ is equal to the area of $\triangle A B C$, find the value of $a$.
15. Let $y=0$ and $x=0$, we get the intersection points of the line $y=-\frac{\sqrt{3}}{3} x+1$ with the $x$-axis and $y$-axis as $A(\sqrt{3}, 0), B(0,1)$, i.e., $O A=\sqrt{3}, O B=1$.
Thus, $A B=2$.
Since $\triangle A B C$ is an isosceles right triangle,
$S_{\triangle A B C}=2$.
Also, since $S_{\triangle A B P}=S_{\triangle A B C}$, thus $S_{\triangle A B P}=2$.
As shown in Figure 7, connect
$P O$.
$S_{\triangle A O P}=\frac{\sqrt{3}}{4}$,
$S_{\triangle B O P}$
$=-\frac{a}{2}$,
$S_{\triangle A O B}=\frac{\sqrt{3}}{2}$.
Since $S_{\triangle B O P}+S_{\triangle \triangle O B}-S_{\triangle M O P}=S_{\triangle A B P}$,
$\therefore-\frac{a}{2}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{4}=2$. Therefore, $a=\frac{\sqrt{3}-8}{2}$