To find the shortest distance between the circle and point P, we need to determine PC, where C is the point on the circle such that PC is perpendicular to the circle's radius OA.

Given:

• PB = 20 cm
• OA = 15 cm

First, we understand that triangle OAP is a right triangle because OA is a radius perpendicular to the tangent PB at point A. Therefore, OP (the hypotenuse) can be calculated using the Pythagorean theorem:

\(OP^2 = OA^2 + AP^2\)

However, instead, we use the relationship of the tangent from a point to a circle:

\(OP^2 = OA^2 + AP^2\) and \(AP = \sqrt{OP^2 - OA^2}\).

Since PB is a tangent, we can assume \(AP=AO\) due to symmetry given \(PB = OP\), the line drawn from the external point is equal as it cuts at circle tangents and through the center PARTS angle at point P.

The correct form is solved as:

\(PC = PB - OA = 20 - 15 = 5 \text{ cm}\)

The shortest distance from the point P to circle \(= PC \text{ cm}\) by figuring distance in-between perpendicular drawn points and circle ends exactly within required nearest distance from the circle.

Thus, the shortest distance is 10 cm.