(1)证明:$\because PA，$$PB$是$\odot O$的切线，$OA，$$OB$是$\odot O$的半径，$\therefore OA\perp PA，$$OB\perp PB.$$\because OC// PA，$$CD\perp AP，$$\therefore CD\perp OC.$$\therefore \angle OAD=\angle CDA=\angle OCD = 90^{\circ}.$$\therefore$四边形$OADC$是矩形.$\therefore OC = AD$
(2)解:设$OC = AD = x.$$\because$四边形$OADC$是矩形，$\odot O$的半径是$3，$$PA = 9，$$\therefore OA = OB = CD = 3，$$PD = PA - AD = 9 - x.$$\because OC// PA，$$\therefore \angle OCB = \angle P.$$\because OB\perp PB，$$CD\perp AP，$$\therefore \angle OBC = \angle CDP = 90^{\circ}.$在$\triangle OCB$和$\triangle CPD$中，$\begin{cases}\angle OBC=\angle CDP = 90^{\circ},\\\angle OCB=\angle P,\\OB = CD,\end{cases}$ $\therefore \triangle OCB\cong\triangle CPD.$$\therefore BC = DP = 9 - x.$在$Rt\triangle OCB$中，由勾股定理，得$OC^{2}=OB^{2}+BC^{2}，$$\therefore x^{2}=3^{2}+(9 - x)^{2}，$解得$x = 5.$$\therefore OC = x = 5$
