Step 1: Find the intersection of the two lines.
The lines are $\sqrt{3}x - y + 2 = 0$ and $\sqrt{3}x + y - 2 = 0$. Adding: $2\sqrt{3}x = 0$, so $x = 0$, then $y = 2$. Intersection is $(0, 2)$.
Step 2: Find the angle each line makes with the x-axis.
Line 1: $y = \sqrt{3}x + 2$ has slope $\sqrt{3}$, so it makes angle $60°$ with x-axis. Line 2: $y = 2 - \sqrt{3}x$ has slope $-\sqrt{3}$, angle $120°$. The angle between them is $60°$, so they are symmetric about the y-axis.
Step 3: Place P on Line 1 at distance 5 from $(0,2)$.
Direction vector of Line 1 is $(\cos 60°, \sin 60°) = (1/2, \sqrt{3}/2)$. So $P = (0 + 5 \cdot \frac{1}{2},\ 2 + 5 \cdot \frac{\sqrt{3}}{2}) = \left(\frac{5}{2},\ 2 + \frac{5\sqrt{3}}{2}\right)$, or the other side $P = \left(-\frac{5}{2},\ 2 - \frac{5\sqrt{3}}{2}\right)$.
Step 4: Find the foot of perpendicular of P onto the y-axis.
The foot of perpendicular from $(x_0, y_0)$ onto the y-axis is $(0, y_0)$. So its distance from origin is $|y_0|$.
Step 5: Compute the distances.
For $P = \left(\frac{5}{2},\ 2 + \frac{5\sqrt{3}}{2}\right)$: distance from origin to foot $= 2 + \frac{5\sqrt{3}}{2}$ (positive, so this is the answer).
Step 6: Verify the other case.
For $P = \left(-\frac{5}{2},\ 2 - \frac{5\sqrt{3}}{2}\right)$: foot is $(0, 2 - \frac{5\sqrt{3}}{2})$, distance $= \left|2 - \frac{5\sqrt{3}}{2}\right|$. Since $\frac{5\sqrt{3}}{2} \approx 4.33 > 2$, this is negative inside. So this equals $\frac{5\sqrt{3}}{2} - 2$. The question asks for distance from origin, so both are valid but the matching option is $2 + \frac{5\sqrt{3}}{2}$.
\[ \boxed{2 + \dfrac{5\sqrt{3}}{2}} \]