is it known that the circle's center lies on the line OP? Is C the center of the circle? If yes, it's pretty obvious that $\displaystyle CP = \cos\theta$.
For the second equality, let A' be the projection of A on OB. Use the fact that the triangle BA'A is similar to the triangle BOD.
Because $\displaystyle \angle OCA=\theta$. The fact that $\displaystyle CP=\cos\theta$ follows from the definition of cosine.
Draw a horizontal line through A and call its intersection with the line OB by A'. Then the triangle BA'A is similar to the triangle BOD because all three angles are equal. Therefore the ratio of A'A = OP to OD is the same as the raio of BA' to BO.and can you explain the 2nd part in more details.