Geometry

$\angle OAP=90^\circ$

$\angle ABP=\angle BAP=\frac12(180^\circ-70^\circ)=55^\circ$

$\angle OAB=\angle OAP -\angle BAP=90^\circ-55^\circ=\boxed{35^\circ}$

Relevant wiki: Tangent to Circles - Problem Solving

$\triangle BAP$ is isosceles since $PA=PB$ (Two tangents drawn from an external point are equal in lengths.)

Thus

$\angle PBA=\angle PAB=\dfrac{180-70}{2}=55^\circ$

Consider quadrilateral $BOAP$ . The sum of interior angles of a quadrilateral is $360^\circ$ .

Hence

$\angle BOA=360-90-90-70=110^\circ$

Since $\triangle BOA$ is isosceles, we have

$\angle OAB=\angle OBA=\dfrac{180-110}{2}=$ $\boxed{35^\circ}$

Note:

The $136^\circ$ is not needed.