Given $y=x^3+ax-b$

$\because$ it passes through $\left(1,-5\right)$

$\therefore 1+a-b=-5$

$\Rightarrow a-b=-6 ...\left(1\right)$

Slope of the line $ax+by+c=0$ is $-\frac{a}{b},$ hence, slope of the line $-x+y+4=0$ is $1$

We know that if two lines are perpendicular, then the product of their slopes is $-1.$

$\therefore$ Slope of tangent of $y,$ which is perpendicular to $-x+y+4=0$ is $-1$

Also, we know that the slop of tangent to a curve $y=f\left(x\right)$ at a point $\left(x_1, y_1\right)$ is given by $f^{\text{'}}\left(x_1\right).$

$\therefore f^{\text{'}}\left(1\right)=-1$

And, $f^{\text{'}}\left(x\right)=3x^2+a$

$\Rightarrow 3+a=-1$

$\Rightarrow a=-4 ...\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get $a=-4, b=2$

$\therefore f\left(x\right)=y=x^3-4x-2$

At, $x=2, y=2^3-4\times 2-2=-2$ and at $x=-2, y={\left(-2\right)}^3-4\times 2-2=-18.$

Thus, out of the given options only $\left(2,-2\right)$ satisfy the curve, hence this point will lie on the curve.