The information given is that a point $P(2ap,ap^2)$ on the parabola $x^2=4ay$. The normal to the parabola at P intersects the parabola again at $Q(2aq,aq^2)$. O is the origin of the graph.
The equation of $PQ$ is $x+py-2ap-ap^3=0$ and the line PQ is a normal to the tangent at P.
Using this information, prove that $p^2+pq+2=0$
Part 2 is to show that $p^2=2$, if the chord $OP$ and $OQ$ are perpendicular.
Use the fact that the product of the slopes of two perpendicular lines is equal to -1
Slope of the line PQ is derived from
$$m_1 =\frac{a(q^2-p^2)}{2a(p-q)} = \frac{(p+q)}{2}$$
Slope of the tangent at $P = \frac{dy}{dx} = \frac{x}{2a}$ evaluated at P which gives
$$ m_2= \frac{2ap}{2a} = p$$
Now $$m_1m_2 = -1 => \frac{p(p+q)}{2} = -1$$
Rearranging you get $p^2+pq+2 =0$
Part II, use similar reasoning, slope of OP $=m_1= \frac{p}{2}$ and that of OQ $m_2= \frac{q}{2}$
$\frac{pq}{4} = -1 => pq = -4$
Substitute this value in the prior equation and get
$p^2 -4+2 = 0 => p^2 = 2$