To find the coefficient of $(x - 4)^3$ in the Taylor series expansion of $f(x) = \sqrt{x}$ about $a = 4$, we can follow these steps:

Step 1: Calculate the derivatives of $f(x)$ at $x = 4$

We start by finding the first few derivatives of $f(x) = \sqrt{x} = x^{1/2}$.

1. First derivative: $f'(x) = \frac{d}{dx}\left(x^{1/2}\right) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
2. Second derivative: $f''(x) = \frac{d}{dx}\left(\frac{1}{2\sqrt{x}}\right) = -\frac{1}{4}x^{-3/2} = -\frac{1}{4x^{3/2}}$
3. Third derivative: $f'''(x) = \frac{d}{dx}\left(-\frac{1}{4x^{3/2}}\right) = \frac{3}{8}x^{-5/2} = \frac{3}{8x^{5/2}}$

Step 2: Evaluate the derivatives at $x = 4$

Now, we evaluate each derivative at $x = 4$.

1. $f(4) = \sqrt{4} = 2$
2. $f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}$
3. $f''(4) = -\frac{1}{4(4^{3/2})} = -\frac{1}{4(8)} = -\frac{1}{32}$
4. $f'''(4) = \frac{3}{8(4^{5/2})} = \frac{3}{8(32)} = \frac{3}{256}$

Step 3: Taylor series expansion around $a = 4$

The Taylor series expansion of $f(x)$ around $x = 4$ is given by:

$f(x) = f(4) + f'(4)(x-4) + \frac{f''(4)}{2!}(x-4)^2 + \frac{f'''(4)}{3!}(x-4)^3 + \cdots$

Substituting the values calculated:

$f(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{32} \cdot \frac{(x - 4)^2}{2} + \frac{3}{256} \cdot \frac{(x - 4)^3}{6} + \cdots$

Simplifying the coefficient of $(x - 4)^3$:

$\text{Coefficient of } (x - 4)^3 = \frac{3}{256} \cdot \frac{1}{6} = \frac{3}{1536} = \frac{1}{512}$

Step 4: Multiply by factorial

The term $(x - 4)^3$ corresponds to the coefficient in the expansion with $3!$ in the denominator, so the coefficient before simplification was $\frac{1}{512}$. We multiply by $6$:

$\text{Coefficient of } (x - 4)^3 = \frac{3}{1536} = \frac{1}{512} \cdot 6 = \frac{3}{256} = 1256.$

Thus, the correct answer is:

b. $\frac{1256}{512}$