The given problem involves a circle where AB is the diameter, and CD is a chord perpendicular to AB intersecting at P. We need to find the radius of the circle using the given information: CP = 2 and PB = 1.

To solve this, we can use the properties of circles and right triangles:

1. Since AB is the diameter, the center of the circle, O, is the midpoint of AB. Thus, AO = OB = radius (r).
2. In the right triangle CPB, by the Pythagorean theorem: \( CP^2 + PB^2 = CB^2 \).
3. Substituting the given values: \( 2^2 + 1^2 = CB^2 \) which simplifies to \( 4 + 1 = CB^2 \), so \( CB = \sqrt{5} \).
4. Since CD is perpendicular to AB and O is the midpoint of AB, OP is the radius.
5. The right triangle OBP gives: \( OP^2 + PB^2 = OB^2 \).
6. We have: \( OP^2 + 1^2 = r^2 \).
7. To find OP, observe the triangle COP: \( CP^2 + OP^2 = CO^2 \).
8. \( 4 + OP^2 = r^2 \).
9. Since OP is a part of a larger radius: \( CO = r\).
10. Equating: \( r^2 - OP^2 = 1 \) and \( r^2 = 4 + OP^2 \), solve them:
11. From step 8 and 9:
    • Let \( OP^2 = x \).
    • Then, \( 4 + x = r^2 \).
    • And \( r^2 - x = 1 \) combining gives:
    • \( 4 + x - x = r^2 - x + x \subset 5 = r^2 \).
    • Therefore, the radius is \( \sqrt{2.5} \).
12. Thus, the radius of the circle is 2.5.

Therefore, the correct answer is 2.5.