6. As shown in Figure 1, the radius of $\odot O$ is $6, M$ is a point outside $\odot O$, and $O M=12$. A line through $M$ intersects $\odot O$ at points $A, B$. The points $A, B$ are symmetric to points $C, D$ with respect to $O M$. $A D$ and $B C$ intersect at point $P$. Then the length of $O P$ is ( )
(A) 4
(B) 3.5
(C) 3
(D) 2.5
6. C.

As shown in Figure 4, extend $M O$ to intersect $\odot O$ at point $N$, and connect $O B$ and $O D$.

By symmetry, point $P$ lies on line $M N$.

Since $\angle B A D=\frac{1}{2} \angle B O D=\angle B O N$, points $A$, $B$, $O$, and $P$ are concyclic.
Thus, $O M \cdot P M=M A \cdot M B=O M^{2}-O N^{2}$
$\Rightarrow P M=O M-\frac{O N^{2}}{O M}=12-3=9$.
Therefore, $O P=O M-P M=3$.