A thin uniform hoop of mass m and radius R rolls without slipping on a horizontal floor with centre speed v and meets a fixed vertical step of height…

During the impulsive contact the only external impulse acts at the step edge, so angular momentum about the edge is conserved. Before impact, the spin part is $I\omega=mR^2(v/R)=mRv$ and the orbital part is $mv(R-R/2)=mvR/2$, giving $L=\tfrac32 mvR$. Afterwards the hoop turns about the edge with $I_e=mR^2+mR^2=2mR^2$, so $\omega'=L/I_e=3v/(4R)$ and the kinetic energy is $\tfrac12 I_e\omega'^2=\tfrac{9}{16}mv^2$. To reach the top the centre must rise by $R/2$, so $\tfrac{9}{16}mv^2\ge mgR/2$ and $v_{min}=\sqrt{8gR/9}$. Option A ignores the collision loss (it equates $mv^2$ to $mgR/2$), which wrongly assumes all kinetic energy survives the impact.