If `P` is a point on the altitude AD of the triangle ABC such the `/_C B P=B/3,` then AP is equal to `2asinC/3` (b) `2bsinC/3` (c) `2csinB/3` (d) `2csinC/3`

Explanation

To solve the problem, we need to find the length of segment \( AP \) in triangle \( ABC \) given that \( P \) is a point on the altitude \( AD \) and \( \angle CBP = \frac{B}{3} \).

### Step-by-Step Solution:

1. **Draw the Triangle**:
- Start by sketching triangle \( ABC \) with \( A \) at the top, \( B \) at the bottom left, and \( C \) at the bottom right. Draw the altitude \( AD \) from \( A \) to line \( BC \), and mark point \( P \) on \( AD \).

2. **Identify Angles**:
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