6. As shown in Figure 1, in the Cartesian coordinate system, the vertex $A$ of the right triangle $\triangle O A B$ is on the positive $x$-axis, $B(3, \sqrt{3}), C\left(\frac{1}{2}, 0\right)$, and $P$ is a moving point on the hypotenuse $O B$. Then the minimum value of $P A+P C$ is
(A) $\frac{\sqrt{13}}{2}$
(B) $\frac{\sqrt{31}}{2}$
(C) $\frac{3+\sqrt{19}}{2}$
(D) $2 \sqrt{7}$
6. B.

As shown in Figure 5, construct the symmetric point $D$ of $A$ with respect to $OB$, connect $CD$, intersecting $OB$ at point $P$, connect $AP$, and draw $DN \perp OA$ at point $N$. At this point, the value of $PA + PC$ is minimized.

Notice that $DP = PA$, then
$$
PA + PC = PD + PC = CD.
$$

It is easy to see that $AB = \sqrt{3}$, $OA = 3$, and $\angle B = 60^\circ$.
By the Pythagorean theorem, $OB = 2\sqrt{3}$.
Thus, $AM = \frac{AO \cdot AB}{OB} = \frac{3}{2}$, and $AD = 2 \times \frac{3}{2} = 3$.
It is easy to get $AN = \frac{1}{2} \times AD = \frac{3}{2}$.
By the Pythagorean theorem,
$DN = \sqrt{3^2 - \left(\frac{3}{2}\right)^2} = \frac{3\sqrt{3}}{2}$.
Given $C\left(\frac{1}{2}, 0\right)$, then $CN = 3 - \frac{1}{2} - \frac{3}{2} = 1$.
In the right triangle $\triangle DNC$, by the Pythagorean theorem,
$DC = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + 1^2} = \frac{\sqrt{31}}{2}$.
Therefore, the minimum value of $PA + PC$ is $\frac{\sqrt{31}}{2}$.