Step 1: $PA$ is tangent at $A$, so $OA \perp PA$, i.e. $\angle OAP = 90^\circ$.
Also $AB \perp OP$, so $AB$ is the altitude from $A$ to the hypotenuse $OP$ of right $\triangle OAP$.
Step 2: In right $\triangle OBA$ (right angled at $B$), by Pythagoras theorem :
$$OB^2 = OA^2 - AB^2 = (10)^2 - (8)^2 = 100 - 64 = 36$$
$$\Rightarrow OB = 6\ \text{cm}$$
Step 3: In a right triangle, the altitude to the hypotenuse satisfies
$$AB^2 = OB \times BP$$
(This follows from $\triangle OBA \sim \triangle ABP$.)
Step 4: $$(8)^2 = 6 \times BP \Rightarrow 64 = 6\,BP$$
Step 5: $$BP = \frac{64}{6} = \frac{32}{3} \approx 10{\cdot}67\ \text{cm}$$
Hence $PB = \dfrac{32}{3}$ cm.