Let $ AB$ be the diameter of a given circle with radius $ 1$ unit, and let $ P$ be a given point on $ AB$. A line through $ P$ meets the circle at points $ C$ and $ D$, so a convex quadrilateral $ ABCD$ is formed. Find the maximum possible area of the quadrilateral.
1. **Setup and Definitions**:
   - Let \( AB \) be the diameter of a circle with radius \( 1 \) unit.
   - Let \( P \) be a point on \( AB \).
   - A line through \( P \) intersects the circle at points \( C \) and \( D \), forming a convex quadrilateral \( ABCD \).

2. **Area Calculation**:
   - The area \( S \) of quadrilateral \( ABCD \) can be calculated using the formula for the area of a cyclic quadrilateral:
     \[
     S = \frac{1}{2} \cdot AB \cdot CD \cdot \sin(\angle APB)
     \]
   - Since \( AB \) is the diameter of the circle, \( AB = 2 \).

3. **Expression for \( CD \)**:
   - Let \( x \) be the perpendicular distance from the center \( O \) of the circle to the line \( CD \).
   - The length \( CD \) can be expressed as:
     \[
     CD = 2 \sqrt{1 - x^2}
     \]

4. **Expression for \( \sin(\angle APB) \)**:
   - Let \( \phi = \angle OPD \).
   - Since \( P \) lies on \( AB \), the angle \( \angle APB \) is the same as \( \angle OPD \).
   - Therefore, \( \sin(\angle APB) = \sin(\phi) = \frac{x}{1} = x \).

5. **Maximizing the Area**:
   - The area \( S \) can now be written as:
     \[
     S = \frac{1}{2} \cdot 2 \cdot 2 \sqrt{1 - x^2} \cdot x = 2x \sqrt{1 - x^2}
     \]
   - To maximize \( S \), we need to maximize the function \( f(x) = x \sqrt{1 - x^2} \).

6. **Finding the Maximum of \( f(x) \)**:
   - Consider the function \( f(x) = x \sqrt{1 - x^2} \).
   - To find the maximum, we take the derivative and set it to zero:
     \[
     f'(x) = \sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}} = \frac{1 - 2x^2}{\sqrt{1 - x^2}}
     \]
     \[
     f'(x) = 0 \implies 1 - 2x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \frac{\sqrt{2}}{2}
     \]

7. **Maximum Area Calculation**:
   - Substituting \( x = \frac{\sqrt{2}}{2} \) into \( f(x) \):
     \[
     f\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2} \sqrt{1 - \left(\frac{\sqrt{2}}{2}\right)^2} = \frac{\sqrt{2}}{2} \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2}
     \]
   - Therefore, the maximum area \( S \) is:
     \[
     S_{\text{max}} = 2 \cdot \frac{1}{2} = 1
     \]

The final answer is \(\boxed{1}\)