To solve this problem, we can use properties of circles and similar triangles. Here’s the step-by-step approach:

1. Understand the configuration:

   • The circle is centered at $A$ with radius 60 units.
   • Points $A$, $C$, $D$, and $P$ lie on a straight line, and points $B$ and $D$ lie on the circle.
   • $\angle CBD \cong \angle DBP$.

2. Identify key angle properties:

   • Since $\angle CBD \cong \angle DBP$, triangle $CBD$ and triangle $DBP$ are similar by the Angle-Angle (AA) similarity criterion.
   • This similarity implies that $\frac{AC}{AP} = \frac{CD}{DP}$ (since similar triangles have proportional sides).

3. Apply the Power of a Point theorem:

   • The Power of a Point theorem states that if a point (here, $C$) lies outside a circle, and two line segments (here, $CD$ and $CP$) are drawn such that they intersect the circle, then $AC \times AP = (radius)^2$.
   • Since $A$ is the center and the radius is 60 units, we have: $AC \times AP = 60^2 = 3600.$

Thus, the product $AC \times AP = 3600$.

Would you like a more detailed breakdown or have any questions?

Here are some related questions to deepen understanding:

1. What is the Power of a Point theorem and how is it applied?
2. How does the Angle-Angle similarity criterion work in triangle similarity?
3. Can you prove why similar triangles have proportional corresponding sides?
4. Why does knowing the radius allow us to find $AC \times AP$ directly?
5. What other geometric configurations can the Power of a Point theorem be applied to?

Tip: Remember that the Power of a Point theorem is a powerful tool in circle geometry, helping to relate distances of points inside or outside a circle to the radius.