过$P$作$PM\bot y$轴于$M$，$PN\bot x$轴于$N$，
$\because P\left(3,3\right)$，
$\therefore PN=PM=3$，
$\because x$轴$\bot y$轴，
$\therefore \angle MON=\angle PNO=\angle PMO=90^{\circ}$，
$\therefore \angle MPN=360^{\circ}-90^{\circ}-90^{\circ}-90^{\circ}=90^{\circ}$，
则四边形$MONP$是正方形，
$\therefore OM=ON=PN=PM=3$，
$\because \angle APB=90^{\circ}$，
$\therefore \angle APB=\angle MON$，
$\therefore \angle MPA=90^{\circ}-\angle APN$，$\angle BPN=90^{\circ}-\angle APN$，
$\therefore \angle APM=\angle BPN$，
在$\triangle APM$和$\triangle BPN$中
$\left\{\begin{array}{l}{∠APM=∠BPN}\\{PM=PN}\\{∠PMA=∠PNB}\end{array}\right.$
$\therefore \triangle APM$≌$\triangle BPN\left(ASA\right)$，
$\therefore AM=BN$，
$\therefore OA+OB$
$=OA+0N+BN$
$=OA+ON+AM$
$=ON+OM$
$=3+3$
$=6$，
故答案为：$6$.

过$P$作$PM\bot y$轴于$M$，$PN\bot x$轴于$N$，
$\because P\left(3,3\right)$，
$\therefore PN=PM=3$，
$\because x$轴$\bot y$轴，
$\therefore \angle MON=\angle PNO=\angle PMO=90^{\circ}$，
$\therefore \angle MPN=360^{\circ}-90^{\circ}-90^{\circ}-90^{\circ}=90^{\circ}$，
则四边形$MONP$是正方形，
$\therefore OM=ON=PN=PM=3$，
$\because \angle APB=90^{\circ}$，
$\therefore \angle APB=\angle MON$，
$\therefore \angle MPA=90^{\circ}-\angle APN$，$\angle BPN=90^{\circ}-\angle APN$，
$\therefore \angle APM=\angle BPN$，
在$\triangle APM$和$\triangle BPN$中
$\left\{\begin{array}{l}{∠APM=∠BPN}\\{PM=PN}\\{∠PMA=∠PNB}\end{array}\right.$
$\therefore \triangle APM$≌$\triangle BPN\left(ASA\right)$，
$\therefore AM=BN$，
$\therefore OA+OB$
$=OA+0N+BN$
$=OA+ON+AM$
$=ON+OM$
$=3+3$
$=6$，
故答案为：$6$.