Let's denote the constant acceleration with which the body moves as $$a$$. We know that the displacement $$S$$ covered by a body starting from rest under a constant acceleration $$a$$ in time $$t$$ is given by the equation of motion: $$S = \frac{1}{2} a t^2$$.

Considering the displacement $$\mathrm{S}_1$$ in first $$(p-1)$$ seconds, we apply the equation of motion:

$$ \mathrm{S}_1 = \frac{1}{2} a (p-1)^2 $$

Similarly, for the displacement $$\mathrm{S}_2$$ in first $$p$$ seconds:

$$ \mathrm{S}_2 = \frac{1}{2} a p^2 $$

To find out the total time it will take to cover the displacement $$\mathrm{S}_1+\mathrm{S}_2$$, we first find the sum of these two displacements:

$$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2 $$

Let's simplify this:

$$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left((p-1)^2 + p^2\right) $$ $$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(p^2 - 2p + 1 + p^2\right) $$ $$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(2p^2 - 2p + 1\right) $$ $$ \mathrm{S}_1+\mathrm{S}_2 = a \left(\frac{2p^2 - 2p + 1}{2}\right) $$

If we consider the total displacement $$\mathrm{S}_1+\mathrm{S}_2$$ is to be covered in a time $$t$$ seconds from rest, we should set this equal to the equation of motion:

$$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a t^2 $$

Equating the two equations:

$$ a \left(\frac{2p^2 - 2p + 1}{2}\right) = \frac{1}{2} a t^2 $$

Since $$a \neq 0$$, we can simplify by dividing both sides by $$ \frac{1}{2} a$$:

$$ \frac{2p^2 - 2p + 1}{2} = \frac{t^2}{2} $$ $$ 2p^2 - 2p + 1 = t^2 $$

To find $$t$$, we take the square root of both sides:

$$ t = \sqrt{2p^2 - 2p + 1} $$

Therefore, the time taken to cover the displacement $$\mathrm{S}_1+\mathrm{S}_2$$ will be:

$$ t = \sqrt{(2p^2 - 2p + 1)}\ s $$

Hence, the correct option would be:

Option D: $$ \sqrt{(2p^2 - 2p + 1)}\ s $$