Using Work-Energy Theorem

$7\mathrm{mgR}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left(\frac{2}{5}{\mathrm{mr}}^2\right)\times {\mathrm{\omega }}^2$

since the sphere is rolling without slipping  $\mathrm{v}=\mathrm{r\omega }$

$\therefore 7\mathrm{mgR}=\frac{1}{2}\times \mathrm{ }\left(\frac{7}{5}{\mathrm{mr}}^2\right)\times \mathrm{ }\frac{{\mathrm{v}}^2}{{\mathrm{r}}^2}$

$\Rightarrow {\mathrm{v}}^2=10\mathrm{gR}$

at point $\mathrm{P}$, we can consider sphere as a point object doing circular motion in a circle of the radius $(\mathrm{R}-\mathrm{r})$ with velocity $\mathrm{v}$,

Hence at point $\mathrm{P}$

$\mathrm{N}=\frac{{\mathrm{mv}}^2}{\left(\mathrm{R}-\mathrm{r}\right)}=\frac{10\mathrm{mgR}}{\left(\mathrm{R}-\frac{\mathrm{R}}{10}\right)}=\frac{10\mathrm{mgR}}{{\displaystyle \frac{9\mathrm{R}}{10}}}$   ( given $\mathrm{r}=\frac{\mathrm{R}}{10}$ )

or $\mathrm{N}=\frac{100}{9}\mathrm{mg}$