To find the angle between $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$, we first need to understand the relationship between force and momentum. The force $$\vec{F}$$ acting on a particle is related to the rate of change of its linear momentum $$\overrightarrow{\mathrm{p}}$$ with respect to time, as described by Newton's second law of motion:

$$\vec{F} = \frac{d\overrightarrow{\mathrm{p}}}{dt}$$

Given the expression for the momentum $$\overrightarrow{\mathrm{p}}(t) = \hat{i} \cos (kt) - \hat{j} \sin (kt)$$, we can find $$\vec{F}$$ by differentiating $$\overrightarrow{\mathrm{p}}$$ with respect to $$t$$:

$$\frac{d\overrightarrow{\mathrm{p}}}{dt} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$$

So, $$\vec{F} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$$.

Now, to find the angle between $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$, we use the dot product formula:

$$\vec{F} \cdot \overrightarrow{\mathrm{p}} = |\vec{F}| |\overrightarrow{\mathrm{p}}| \cos(\theta)$$,

where $$\theta$$ is the angle between $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$. However, in this case, it's more insightful to see if $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$ are orthogonal (at a $$\frac{\pi}{2}$$ angle to each other), because the dot product of two perpendicular vectors is zero.

The dot product of $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$ is:

$$(-k\hat{i} \sin (kt) - k\hat{j} \cos (kt)) \cdot (\hat{i} \cos (kt) - \hat{j} \sin (kt)) =$$

$$- k \sin (kt) \cos (kt) + k \cos (kt) \sin (kt) = 0$$

The result is zero, indicating that the angle between $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$ is indeed $$\frac{\pi}{2}$$.

Therefore, the correct option is:

Option A $$\frac{\pi}{2}$$.