Given $y=4x^4+x$

Differentiating the above equation,

$\Rightarrow \frac{dy}{dx}=16x^3+1$

The tangent at point $(0, 0)$ has a slope $=16\left(0\right)+1=1$

The tangent at point $P(a,b)$ has a slope $16a^3+1$

$m_1{m}_2=-1$ when the lines with slope $m_1$ and $m_2$ are perpendicular to each other.

$\therefore \left(16a^3+1\right)\left(1\right)=-1$

$\Rightarrow a^3=-\frac{1}{8}$

$\Rightarrow a=\frac{-1}{2}$

Point $P(a, b)$ lies on curve $\Rightarrow b=4{\left(a\right)}^4+a$

$\Rightarrow b=\frac{-1}{4}$

Hence, $P\left(\frac{-1}{2},\frac{-1}{4}\right)$