$ABCD$ is a rectangle with $AB = CD = 2$. A circle centered at $O$ is tangent to $BC$, $CD$, and $AD$ (and hence has radius $1$). Another circle, centered at $P$, is tangent to circle $O$ at point $T$ and is also tangent to $AB$ and $BC$. If line $AT$ is tangent to both circles at $T$, find the radius of circle $P$.
1. **Identify the given information and set up the problem:**
   - Rectangle \(ABCD\) with \(AB = CD = 2\).
   - Circle centered at \(O\) is tangent to \(BC\), \(CD\), and \(AD\) with radius \(1\).
   - Another circle centered at \(P\) is tangent to circle \(O\) at point \(T\) and is also tangent to \(AB\) and \(BC\).
   - Line \(AT\) is tangent to both circles at \(T\).

2. **Determine the coordinates and positions:**
   - Place the rectangle \(ABCD\) in the coordinate plane with \(A = (0, 0)\), \(B = (2, 0)\), \(C = (2, 2)\), and \(D = (0, 2)\).
   - The circle centered at \(O\) is tangent to \(BC\), \(CD\), and \(AD\), so its center \(O\) is at \((1, 1)\) with radius \(1\).

3. **Analyze the second circle centered at \(P\):**
   - The circle centered at \(P\) is tangent to circle \(O\) at point \(T\) and is also tangent to \(AB\) and \(BC\).
   - Let the radius of the circle centered at \(P\) be \(r\).

4. **Determine the coordinates of \(P\):**
   - Since the circle is tangent to \(AB\) and \(BC\), the center \(P\) must be at \((2 - r, r)\).

5. **Use the tangency condition at point \(T\):**
   - The distance between the centers \(O\) and \(P\) is \(1 + r\) (sum of the radii).
   - The distance between \(O\) at \((1, 1)\) and \(P\) at \((2 - r, r)\) is given by:
     \[
     \sqrt{(2 - r - 1)^2 + (r - 1)^2} = \sqrt{(1 - r)^2 + (r - 1)^2} = \sqrt{2(1 - r)^2} = \sqrt{2} |1 - r|
     \]
   - Since \(1 - r\) is positive (as \(r < 1\)), we have:
     \[
     \sqrt{2} (1 - r) = 1 + r
     \]

6. **Solve the equation:**
   \[
   \sqrt{2} (1 - r) = 1 + r
   \]
   \[
   \sqrt{2} - \sqrt{2} r = 1 + r
   \]
   \[
   \sqrt{2} - 1 = (1 + \sqrt{2}) r
   \]
   \[
   r = \frac{\sqrt{2} - 1}{1 + \sqrt{2}}
   \]

7. **Rationalize the denominator:**
   \[
   r = \frac{\sqrt{2} - 1}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{(\sqrt{2} - 1)(1 - \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})}
   \]
   \[
   r = \frac{\sqrt{2} - 2 - 1 + \sqrt{2}}{1 - 2} = \frac{2\sqrt{2} - 3}{-1} = 3 - 2\sqrt{2}
   \]

The final answer is \(\boxed{3 - 2\sqrt{2}}\).