The number of distinct real roots of $$\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$$ in the interval $$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$$ is:

We need to find the number of distinct real roots of the equation:

$$\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$$

in the interval $$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$$.

Let us expand this determinant. We apply the column operation $$C_1 \to C_1 + C_2 + C_3$$:

$$C_1$$ becomes: $$\sin x + \cos x + \cos x = \sin x + 2\cos x$$ for each row.

So we can factor out $$(\sin x + 2\cos x)$$ from $$C_1$$:

$$(\sin x + 2\cos x) \begin{vmatrix} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{vmatrix} = 0$$

Now we expand the remaining $$3 \times 3$$ determinant. Apply $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$:

$$\begin{vmatrix} 1 & \cos x & \cos x \\ 0 & \sin x - \cos x & 0 \\ 0 & 0 & \sin x - \cos x \end{vmatrix}$$

This is an upper triangular determinant, so its value is the product of diagonal entries:

$$1 \times (\sin x - \cos x) \times (\sin x - \cos x) = (\sin x - \cos x)^2$$

So the original equation becomes:

$$(\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$$

This gives us two cases.

Case 1: $$\sin x - \cos x = 0$$

$$\sin x = \cos x$$

$$\tan x = 1$$

$$x = \frac{\pi}{4}$$

We check: $$x = \frac{\pi}{4}$$ lies in the interval $$\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$$, so this is a valid root.

Case 2: $$\sin x + 2\cos x = 0$$

$$\sin x = -2\cos x$$

$$\tan x = -2$$

We need to check if $$\tan x = -2$$ has a solution in $$\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$$.

In this interval, $$\tan x$$ ranges from $$\tan\left(-\frac{\pi}{4}\right) = -1$$ to $$\tan\left(\frac{\pi}{4}\right) = 1$$.

Since $$-2 < -1$$, the value $$\tan x = -2$$ is outside the range $$[-1, 1]$$. Therefore, there is no solution for this case in the given interval.

Thus, the only root in the interval $$\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$$ is $$x = \frac{\pi}{4}$$.

The number of distinct real roots is $$1$$, which is Option B.