Let the equation of the line having intercepts $a$ and $b$ on the x-axis and y-axis respectively be given by the intercept form:

$$\frac{x}{a} + \frac{y}{b} = 1$$

Rewriting this equation in the general form $Ax + By + C = 0$, we multiply by $ab$ to get:

$$bx + ay - ab = 0$$

Here, the coefficients are $A = b$, $B = a$, and $C = -ab$.

The length of the perpendicular from the origin $(0, 0)$ to this line is given as $p$. Using the perpendicular distance formula from the context, the distance $d$ (or $p$) from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:

$$p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substituting the coordinates of the origin $(0, 0)$ for $(x_1, y_1)$ and the values of $A, B, C$:

$$p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}}$$ $$p = \frac{|-ab|}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}}$$

Squaring both sides of the equation, we get:

$$p^2 = \frac{a^2b^2}{a^2 + b^2}$$

Taking the reciprocal of both sides:

$$\frac{1}{p^2} = \frac{a^2 + b^2}{a^2b^2}$$

Splitting the fraction on the right-hand side:

$$\frac{1}{p^2} = \frac{a^2}{a^2b^2} + \frac{b^2}{a^2b^2}$$

Simplifying the terms:

$$\frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2}$$

Rearranging the terms to match the required result:

$$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$$

∴ Hence proved.