If the common tangents to the parabola, $$x^2 = 4y$$ and the circle, $$x^2 + y^2 = 4$$ intersect at the point $$P$$, then the distance of $$P$$ from the origin (units), is:

We begin with the standard parabola equation $$x^2 = 4y$$. For a parabola of the form $$x^2 = 4ay$$ the slope form of a tangent is

$$y = mx + \frac{a}{m},$$

where $$m$$ is the slope of the tangent and $$a$$ is the parameter of the parabola. In the given parabola $$4a = 4$$, so $$a = 1$$. Substituting $$a = 1$$, every tangent to the parabola can be written as

$$y = mx + \frac{1}{m}\qquad\text{(1)}$$

Next we want the same line to be tangent to the circle $$x^2 + y^2 = 4$$. A line in the form $$y = mx + c$$ can be rewritten as $$mx - y + c = 0$$, and its perpendicular distance from the origin $$(0,0)$$ is given by the formula

$$\text{Distance} = \frac{|c|}{\sqrt{m^2 + 1}}.$$

For tangency to the circle, this distance must equal the radius of the circle. The circle has radius $$2$$. Hence we must satisfy

$$\frac{|c|}{\sqrt{m^2 + 1}} = 2.$$

From equation (1) we identify $$c = \dfrac{1}{m}$$, and so

$$\frac{\left|\dfrac{1}{m}\right|}{\sqrt{m^2 + 1}} = 2.$$

Removing the absolute value (because we will square both sides) and squaring, we have

$$\frac{1}{m^2}\cdot\frac{1}{m^2 + 1} = 4.$$

Multiplying both sides by $$m^2(m^2 + 1)$$ gives

$$1 = 4m^2(m^2 + 1).$$

Expanding the right-hand side,

$$1 = 4m^4 + 4m^2.$$

Bringing all terms to one side,

$$4m^4 + 4m^2 - 1 = 0.$$

Letting $$t = m^2$$ turns this into a quadratic in $$t$$:

$$4t^2 + 4t - 1 = 0.$$

Using the quadratic formula

$$t = \frac{-4 \pm \sqrt{(4)^2 - 4\cdot4\cdot(-1)}}{2\cdot4} = \frac{-4 \pm \sqrt{16 + 16}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} = \frac{-1 \pm \sqrt{2}}{2}.$$

Since $$t = m^2 \ge 0$$, we keep the positive root:

$$m^2 = \frac{\sqrt{2} - 1}{2}.$$

Therefore there are exactly two slopes,

$$m_1 = +\sqrt{\frac{\sqrt{2} - 1}{2}}, \qquad m_2 = -\sqrt{\frac{\sqrt{2} - 1}{2}}.$$

These two slopes give the two common tangents to both the parabola and the circle:

$$\begin{aligned} \text{(i)}\; & y = m_1x + \frac{1}{m_1},\\[4pt] \text{(ii)}\; & y = m_2x + \frac{1}{m_2}. \end{aligned}$$

We now find their point of intersection $$P(x_P,\,y_P)$$. Because the two tangents meet at this point, their $$y$$-values are equal:

$$m_1x_P + \frac{1}{m_1} = m_2x_P + \frac{1}{m_2}.$$

Substituting $$m_2 = -m_1$$ and $$\dfrac{1}{m_2} = -\dfrac{1}{m_1}$$ gives

$$m_1x_P + \frac{1}{m_1} = -m_1x_P - \frac{1}{m_1}.$$

Collecting like terms,

$$m_1x_P + m_1x_P = -\frac{1}{m_1} - \frac{1}{m_1},$$

$$2m_1x_P = -\frac{2}{m_1}.$$

Dividing by $$2m_1$$ yields the $$x$$-coordinate of $$P$$:

$$x_P = -\frac{1}{m_1^2}.$$

But $$m_1^2 = \dfrac{\sqrt{2} - 1}{2}$$, so

$$x_P = -\,\frac{1}{\dfrac{\sqrt{2} - 1}{2}} = -\frac{2}{\sqrt{2} - 1}.$$

To rationalise the denominator we multiply numerator and denominator by $$\sqrt{2} + 1$$:

$$x_P = -\frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = -\frac{2(\sqrt{2} + 1)}{2 - 1} = -2(\sqrt{2} + 1).$$

To find $$y_P$$ we substitute $$x_P$$ in, say, the first tangent:

$$y_P = m_1x_P + \frac{1}{m_1} = m_1\!\Bigl(-\frac{1}{m_1^2}\Bigr) + \frac{1}{m_1} = -\frac{1}{m_1} + \frac{1}{m_1} = 0.$$

Thus the point of intersection is

$$P\bigl(-2(\sqrt{2} + 1),\,0\bigr).$$

The distance of $$P$$ from the origin is simply the absolute value of its $$x$$-coordinate (because $$y_P = 0$$):

$$OP = \left| -2(\sqrt{2} + 1) \right| = 2(\sqrt{2} + 1).$$

Hence, the correct answer is Option D.