If the tangent at (1, 1) on $y^2=x(2-x)^2$ meets the curve again at P, then P is

(9/4, 3/8)

We have,

$y^2=x(2-x)^2$             ........(i)

$\Rightarrow y^2=x^3-4 x^2+4 x$

$\Rightarrow 2 y \frac{d y}{d x}=3 x^2-8 x+4$

$\Rightarrow \frac{d y}{d x}=\frac{3 x^2-8 x+4}{2 y} \Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{3-8+4}{2}=\frac{-1}{2}$

The equation of the tangent at (1, 1) is

$y-1=\frac{-1}{2}(x-1) \Rightarrow x+2 y-3=0$                 ......(ii)

Solving (i) and (ii), we get $x=9 / 4$ and $y=3 / 8$.

Hence, the coordinates of P are $(9 / 4,3 / 8)$.