A

Step 1: Identify the forces acting on the bob when it is at the lowest point. The tension in the rod (T) is twice the weight (W) of the bob, so we have:

$$ T = 2W $$

Since the bob is in circular motion, we can apply Newton's second law in the radial direction at the lowest point:

$$ T - W = \frac{mv^2}{L} $$
where m is the mass of the bob, v is the velocity at the lowest point, and L is the length of the pendulum.

Step 2: Substitute the known values into the equation:

$$ 2W - W = \frac{mv^2}{L} $$

Simplifying gives:

$$ W = \frac{mv^2}{L} $$

Rearranging gives:

$$ v^2 = \frac{WL}{m} $$

Since weight is given by W = mg:

$$ v^2 = \frac{mgL}{m} = gL $$

So, the velocity at the lowest point is:

$$ v = \sqrt{gL} $$

Step 3: Now, consider the maximum height (h) the bob reaches when it swings up to an angle α. At this point, all kinetic energy has been converted to potential energy:

$$ mgh = \frac{1}{2} mv^2 $$

Substituting for v gives:

$$ mg h = \frac{1}{2} m g L $$

Simplifying leads to:

$$ h = \frac{L}{2} $$

Step 4: Use trigonometry to relate height to angle α:
In a right triangle, we have:

$$ h = L (1 - cos(\alpha)) $$

Equating the expressions for height gives:

$$ \frac{L}{2} = L (1 - cos(\alpha)) $$

Dividing both sides by L we get:

$$ \frac{1}{2} = 1 - cos(\alpha) $$

Therefore:

$$ cos(\alpha) = \frac{1}{2} $$

Step 5: Solving for angle α:

$$ \alpha = 60^{\circ} $$

The maximum angle through which the pendulum is deflected is therefore 60 degrees.
Therefore, the answer is:
A.