Solution:

$\boxed{\dfrac{1}{6}}$

We translate the problem into one about 2-D geometry. Consider the right triangle $ABO$, and $P$ is some point on $AO$. Then, the choice of $P$ minimizes $PA + 6PB$. Construct the line $\ell$ through $A$ but outside the triangle $ABO$ so that $\sin \angle(AO, \ell) = \dfrac{1}{6}$. For whichever $P$ chosen, let $Q$ be the projection of $P$ onto $\ell$, then $PQ = \dfrac{1}{6} AP$. Then, since $PA + 6PB = 6(PQ + PB)$, it is equivalent to minimize $PQ + PB$. Observe that this sum is minimized when $B, P, Q$ are collinear and the line through them is perpendicular to $\ell$ (so that $PQ + PB$ is simply the distance from $B$ to $\ell$). Then, $\angle AQB = 90^\circ$, and since $\angle AOB = 90^\circ$ as well, we see that $A, Q, P, B$ are concyclic. Therefore, $\angle PBO = \angle OPA = \angle(AO, \ell)$, and the sine of this angle is therefore $\dfrac{1}{6}$.