In a triangle ABC, $$AB = AC = 8$$ cm. A circle drawn with $$BC$$ as diameter passes through $$A$$. Another circle drawn with center at $$A$$ passes through $$B$$ and $$C$$. Then the area, in sq. cm, of the overlapping region between the two circles is:

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#### Step 1: Key Observations and Parameters  
- **Triangle ABC:** Given that $$AB = AC = 8$$ cm and $$BC$$ is the diameter of the circle passing through $$A$$, we have:  
  - $$\angle BAC = 90^\circ$$ (since the angle subtended by the diameter in a semicircle is a right angle).  
  - Using Pythagoras theorem:  
    $$
    BC = \sqrt{AB^2 + AC^2} = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = 8\sqrt{2}.
    $$

#### Step 2: Radii of the Circles  
- **Circle with center at A (Circle C1):** Radius = $$AB = AC = 8$$ cm.  
- **Circle with diameter BC (Circle C2):** Radius = $$\frac{BC}{2} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$$ cm.

#### Step 3: Overlapping Area Calculation  
The overlapping area consists of two parts:  
1. **Half of Circle C2 (semicircle):**  
   Area of semicircle:  
   $$
   \frac{1}{2} 	imes \pi 	imes (4\sqrt{2})^2 = \frac{1}{2} 	imes \pi 	imes 32 = 16\pi.
   $$

2. **Minor segment created by chord BC in Circle C1:**  
   - Area of sector formed by chord BC in Circle C1:  
     $$
     	ext{Sector Area} = \frac{\angle BAC}{360^\circ} 	imes \pi 	imes r^2 = \frac{90^\circ}{360^\circ} 	imes \pi 	imes (8)^2 = \frac{1}{4} 	imes \pi 	imes 64 = 16\pi.
     $$  
   - Area of triangle ABC:  
     $$
     	ext{Triangle Area} = \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes 8 	imes 8 = 32.
     $$  
   - Minor segment area:  
     $$
     	ext{Segment Area} = (	ext{Sector Area}) - (	ext{Triangle Area}) = 16\pi - 32.
     $$

#### Step 4: Total Overlapping Area  
The total overlapping area is:  
$$
	ext{Total Overlapping Area} = (	ext{Semicircle Area}) + (	ext{Minor Segment Area}) = 16\pi + (16\pi - 32) = 32\pi - 32.
$$

Factorize:  
$$
	ext{Total Overlapping Area} = 32(\pi - 1).
$$

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### Final Answer:  
The area of the overlapping region between the two circles is **$$32(\pi - 1)$$**, and the correct answer is **B**.

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