A block of mass m attached to an ideal spring of force constant k executes simple harmonic motion with amplitude A on a smooth horizontal table. At…

Just before the putty lands, $x=A/2$ and $v=\omega\sqrt{A^2-A^2/4}=\tfrac{\sqrt3}{2}\omega A$ with $\omega^2=k/m$. The putty falls vertically, so no horizontal impulse acts on the block and horizontal momentum is conserved: $mv=2mv'$, hence $v'=v/2$. The position is unchanged, but the angular frequency becomes $\omega'^2=k/2m=\omega^2/2$. The new amplitude follows from $A'^2=x^2+\dfrac{v'^2}{\omega'^2}=\dfrac{A^2}{4}+\dfrac{\tfrac{3}{16}\omega^2A^2}{\omega^2/2}=\dfrac{A^2}{4}+\dfrac{3A^2}{8}=\dfrac{5A^2}{8}$. So $A'=\sqrt{5/8}\,A\approx0.79A$. Option B comes from dropping the position term in the amplitude formula, option C from assuming the putty is dropped at the mean position, and option D from using the old $\omega$ with the reduced speed ($A'^2=A^2/4+3A^2/16=7A^2/16$).