Given:
ABCD is a parallelogram
P is any point on CD
ar(DPA) = 15 cm^2
ar(APC) = 20 cm^2

To Find:
ar(APB)

Solution:
We are given that ABCD is a parallelogram. Therefore, opposite sides of the parallelogram are equal in length and parallel to each other.

Step 1: Identify relevant triangles
In order to find ar(APB), we need to identify the relevant triangles in the parallelogram.

In the parallelogram ABCD, we have the following triangles:
1. Triangle DPA
2. Triangle APC
3. Triangle APB

Step 2: Use the given areas to find ratios
We are given the areas of triangles DPA and APC. We can use these areas to find the ratio of their heights (or lengths of the perpendiculars) with respect to the base CD.

Let the height of triangle DPA be h1 and the height of triangle APC be h2.
Given:
ar(DPA) = 15 cm^2
ar(APC) = 20 cm^2

We know that the area of a triangle is given by the formula: area = (1/2) * base * height.
Using this formula, we can write the following equations:
(1/2) * CD * h1 = 15
(1/2) * CD * h2 = 20

Dividing the second equation by the first equation, we get:
(h2/h1) = (20/15)
Simplifying, we get:
(h2/h1) = (4/3)

Step 3: Use the ratios to find ar(APB)
In the parallelogram ABCD, we can see that triangles DPA and APC share the same base CD.

Since the heights of triangles DPA and APC are in the ratio (4/3), the areas of these triangles will also be in the same ratio.

Therefore, we can write:
ar(DPA)/ar(APC) = (15/20)

Substituting the given values, we get:
15/ar(APC) = (15/20)

Cross-multiplying, we get:
ar(APC) = 20

Since the area of triangle APC is given as 20 cm^2, we can conclude that the area of triangle APB will also be 20 cm^2.

Answer:
ar(APB) = 20 cm^2