$ ABCD$ is a rectangle of area 2.  $ P$ is a point on side $ CD$ and $ Q$ is the point where the incircle of $ \triangle PAB$ touches the side $ AB$.  The product $ PA \cdot PB$ varies as $ ABCD$ and $ P$ vary.  When $ PA \cdot PB$ attains its minimum value, 

a) Prove that $ AB \geq 2BC$,
b) Find the value of $ AQ \cdot BQ$.
### Part (a): Prove that \( AB \geq 2BC \)

1. **Given**: \(ABCD\) is a rectangle with area 2, \(P\) is a point on side \(CD\), and \(Q\) is the point where the incircle of \(\triangle PAB\) touches the side \(AB\).

2. **Objective**: Prove that \(AB \geq 2BC\).

3. **Approach**: We need to find the minimum value of \(PA \cdot PB\). We know that \(PA \cdot PB\) is minimized when \(\sin \angle APB\) is maximized, which occurs when \(\angle APB\) is a right angle.

4. **Calculation**:
   - The area of rectangle \(ABCD\) is given by \(AB \cdot BC = 2\).
   - When \(\angle APB = 90^\circ\), \(PA\) and \(PB\) are the legs of a right triangle with hypotenuse \(AB\).
   - By the Pythagorean theorem, \(PA \cdot PB = AB \cdot BC\).

5. **Condition**:
   - For \(\angle APB\) to be a right angle, the circle with diameter \(AB\) must touch or intersect \(CD\).
   - This implies that the diameter \(AB\) must be at least twice the height \(BC\) of the rectangle, i.e., \(AB \geq 2BC\).

Thus, we have shown that \(AB \geq 2BC\).

### Part (b): Find the value of \(AQ \cdot BQ\)

1. **Given**: \(Q\) is the point where the incircle of \(\triangle PAB\) touches \(AB\).

2. **Objective**: Find the value of \(AQ \cdot BQ\).

3. **Approach**: Use the properties of the incircle and the right triangle \(\triangle PAB\).

4. **Calculation**:
   - The incircle of \(\triangle PAB\) touches \(AB\) at \(Q\).
   - The lengths \(AQ\) and \(BQ\) are given by the segments formed by the tangents from \(A\) and \(B\) to the incircle.
   - For a right triangle, the product of the segments formed by the tangents from the vertices to the incircle is given by:
     \[
     AQ \cdot BQ = \frac{1}{4}(AB + PA - PA)(AB + PB - PA)
     \]
   - Simplifying, we get:
     \[
     AQ \cdot BQ = \frac{1}{4}(AB^2 - (PA - PB)^2)
     \]
   - Since \(PA \cdot PB = AB \cdot BC = 2\) and \(\angle APB = 90^\circ\), we have:
     \[
     AQ \cdot BQ = \frac{1}{2} PA \cdot PB = \frac{1}{2} \cdot 2 = 1
     \]

Thus, the value of \(AQ \cdot BQ\) is 1.

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