First of all let us see what will be PS*PS'.

PS = e*PM, where PM is the distance of P from the diretrix.

PS' = e*PM'. If co-ordinates of P are (x1, y1),

PM = (a/e - x1) and PM' = (a/e + x1). Hence

PS*PS' = e^2(a^2/e^2 - x1^2) = (a^2 - e^2*x1^2)........(1)

The equation of the tangent to the ellipse is given by

xx1/a^2 + yy1/b^2 = 1

Slope of the tangent is -x1*b^2/y1*a^2

Slope of the normal is y1a^2/x1b^2

Find the equation of the normal.

Find the point of intersection of the normal with x-axis.

Find PG^2.

Replace y1^2 in the above expression by b^2(1 - x^2/a^2)

Now simplify.