Slope of Normal at Parameter $$t$$

The given parabola is $$y^{2}=8x$$.
Rewrite it in the standard form $$y^{2}=4ax$$ to identify $$a$$:

$$4a=8 \;\Longrightarrow\; a=2$$

For $$y^{2}=4ax$$ a convenient parametric point is
$$P\,(x,y)=\bigl(a\,t^{2},\,2a\,t\bigr)=\bigl(2t^{2},\,4t\bigr)$$ with $$t\gt 0$$ (first quadrant).

Normal at $$P$$
Slope of the tangent at parameter $$t$$ is $$\dfrac{1}{t}$$, so the slope of the normal is $$-t$$.
Equation of the normal through $$P(2t^{2},4t)$$ is

$$y-4t=-t\bigl(x-2t^{2}\bigr) \;\;\Longrightarrow\;\; y=-tx+4t+2t^{3} \;-(1)$$

The second intersection of this normal with the parabola (besides $$P$$) is $$Q$$.
Substitute $$y$$ from $$(1)$$ into $$y^{2}=8x$$ to locate $$Q$$:

$$( -t x + 4t + 2t^{3})^{2}=8x$$

This is a quadratic in $$x$$ whose two roots are $$x_{P}=2t^{2}$$ and $$x_{Q}$$.
Using the sum of roots formula $$\dfrac{-b}{a}$$:

$$x_{P}+x_{Q}=8+4t^{2}+\dfrac{8}{t^{2}} \;\;\Longrightarrow\;\; x_{Q}=8+2t^{2}+\dfrac{8}{t^{2}} \;-(2)$$

Coordinates of $$Q$$
Insert $$x_{Q}$$ from $$(2)$$ into the normal $$(1)$$:

$$y_{Q}= -t\!\left(8+2t^{2}+\dfrac{8}{t^{2}}\right)+4t+2t^{3} =-4t-\dfrac{8}{t} \;-(3)$$

Right-angle condition at the focus
Focus of the parabola is $$S(2,0)$$.
Vectors $$\overrightarrow{SP}$$ and $$\overrightarrow{SQ}$$ must be perpendicular, i.e. their dot product is zero:

$$\overrightarrow{SP}=(2t^{2}-2,\,4t)$$
$$\overrightarrow{SQ}=\bigl(x_{Q}-2,\,y_{Q}\bigr) =\left(6+2t^{2}+\dfrac{8}{t^{2}},\,-4t-\dfrac{8}{t}\right)$$

Dot product:

$$(2t^{2}-2)\!\left(6+2t^{2}+\dfrac{8}{t^{2}}\right)+4t\!\left(-4t-\dfrac{8}{t}\right)=0$$

Simplifying gives

$$4t^{6}-8t^{4}-28t^{2}-16=0$$

Divide by 4 and put $$u=t^{2}$$ (since $$t\gt0$$):

$$u^{3}-2u^{2}-7u-4=0$$

This factors as $$\bigl(u-4\bigr)\bigl(u+1\bigr)^{2}=0$$, so the positive solution is

$$u=4\;\Longrightarrow\; t^{2}=4\;\Longrightarrow\; t=2$$

Required $$x$$-coordinate of $$P$$

$$x_{P}=2t^{2}=2(4)=8$$

Hence, the point $$P$$ has $$x$$-coordinate $$8$$.

Option A which is: $$8$$