Step-by-Step Solution:

1. Identify the equations of the tangent and the normal. The parabola is given by the equation ${y^2} = 4ax$. The point $P$ is given by $(a{t^2},2at)$. The equation of the tangent to the parabola at a point with parameter 't' is: $ty = x + at^2 \quad \cdots (1)$ The point $S$ is given by $(a{s^2},2as)$. The equation of the normal to the parabola at a point with parameter 's' is: $y - 2as = -s(x - as^2)$ $y + sx = 2as + as^3 \quad \cdots (2)$

2. Use the given condition relating parameters s and t. We are given that $st=1$, which implies $s = 1/t$.

3. Solve the system of equations for the ordinate (y-coordinate) of the intersection point. We have two linear equations for the intersection point $(x, y)$: (1) $ty - x = at^2$ (2) $y + sx = 2as + as^3$

   From equation (1), we can express $x$ in terms of $y$ and $t$: $x = ty - at^2$ Substitute this expression for $x$ into equation (2): $y + s(ty - at^2) = 2as + as^3$ Expanding the left side: $y + sty - ast^2 = 2as + as^3$ Group the terms containing $y$: $y(1 + st) = ast^2 + 2as + as^3$

4. Apply the condition $st=1$ to simplify the equation. Substitute $st=1$ into the left side of the equation: $y(1 + 1) = ast^2 + 2as + as^3$ $2y = a(st)t + 2as + as^3$ Since $st=1$, we have: $2y = a(1)t + 2as + as^3$ $2y = at + 2as + as^3$

5. Express the right side entirely in terms of 't'. Now, substitute $s = 1/t$ into the right side of the equation: $2y = at + 2a(1/t) + a(1/t)^3$ $2y = at + \frac{2a}{t} + \frac{a}{t^3}$

6. Simplify the expression to find the final ordinate. Factor out 'a' and find a common denominator on the right side: $2y = a \left( t + \frac{2}{t} + \frac{1}{t^3} \right)$ $2y = a \left( \frac{t^4 + 2t^2 + 1}{t^3} \right)$ The numerator is a perfect square: $t^4 + 2t^2 + 1 = (t^2+1)^2$. $2y = a \frac{(t^2+1)^2}{t^3}$ Finally, solve for $y$: $y = \frac{a(t^2+1)^2}{2t^3}$

Note on the other given information: The problem provides details about points $Q$, $R$, and $K$, and states that $PQ$ is a focal chord and $QR$ is parallel to $PK$. This information can be used to establish a relationship between the parameters, for instance, $r = (t^2-1)/t$. However, this relationship is not needed to find the ordinate of the intersection point of the tangent at $P$ and the normal at $S$, because the crucial condition $st=1$ is given directly.

Comparing our result with the given options:

A: ${{{{\left( {{t^2} + 1} \right)}^2}} \over {2{t^3}}}$ (missing factor 'a')

B: ${{a{{\left( {{t^2} + 1} \right)}^2}} \over {2{t^3}}}$ (matches our result)

C: ${{a{{\left( {{t^2} + 1} \right)}^2}} \over {{t^3}}}$ (missing factor of 2 in denominator)

D: ${{a{{\left( {{t^2} + 2} \right)}^2}} \over {{t^3}}}$ (numerator is incorrect)

The correct option is B.