If the tangent to the curve,

$y=x^{3}+a x-b$

Since, the point $(1,-5)$ lies on the curve.

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

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

Now, $\frac{d y}{d x}=3 x^{2}+a$

$\Rightarrow\left(\frac{d y}{d x}\right)_{\text {at } x=1}=3+a$

Since, required line is perpendicular to $y=x-4$,

then slope of tangent at the point $P(1,-5)=-1$

$\therefore 3+a=-1$

$\Rightarrow \mathrm{a}=-4$

$\Rightarrow \mathrm{b}=2$

$\therefore$ the equation of the curve is $\mathrm{y}=\mathrm{x}^{3}-4 \mathrm{x}-2$

$\Rightarrow(2,-2)$ lies on the curve