3 Draw a line through the focus $F$ of the parabola $y^{2}=8(x+2)$ with an inclination angle of $60^{\circ}$. If this line intersects the parabola at points $A, B$, the perpendicular bisector of chord $A B$ intersects the x-axis at point $P$, then the length of segment $P F$ is $\mathrm{J}($.
(A) $\frac{16}{3}$
(B) $\frac{8}{3}$
(C) $\frac{16 \sqrt{3}}{3}$
(I)) $8 \sqrt{3}$
From the given conditions, we know that $F$ is the origin of coordinates, so the equation of the line on which the chord $AB$ lies is $y=\sqrt{3} x$. Substituting into the equation of the parabola, we get
$$
3 x^{2}-8 x-16=0
$$

Let the midpoint of $AB$ be $E$. Then the x-coordinate of $E$ is $\frac{4}{3}$. Given that the inclination angle of line $AB$ is $60^{\circ}$, we know $|FE|=2 \times \frac{4}{3}=\frac{8}{3},|PF|=2|FE|=\frac{16}{3}$, so the answer is (A).