Solution:
$12 + 4 \sqrt{3} + \pi$

The shortest path consists of a tangent to the circle, a circular arc, and then another tangent. The first tangent, from $(8 \sqrt{3}, 0)$ to the circle, has length $4 \sqrt{3}$, because it is a leg of a $30$-$60$-$90$ right triangle. The $15^{\circ}$ arc has length $\frac{15}{360}(24\pi)$, or $\pi$, and the final tangent, to $(0,12 \sqrt{2})$, has length $12$.