(a) Let us consider the given diagram the stone is dropped from point $p$ when the stone drops through a length $L$ it will be in free fall. After that the elasticity of the string will force it to a SHM. Let the stone come to rest instantaneously at $y$.
So the change in P.E. of stone at $Q$ and $P$ converts into mechanical energy in string of spring constant $K$
PE of stone = mechanical $E$ of string
$$
\begin{aligned}
&m g y=\frac{1}{2} k(y-L)^2=\frac{1}{2} K\left(y^2+L^2-2 y L\right) \\
&\Rightarrow m g y=\frac{1}{2} k y^2-k y L+\frac{1}{2} k L^2 \\
&\Rightarrow \frac{1}{2} k y^2-(k L+m g) y+\frac{1}{2} k L^2=0 \\
&y=\frac{(k L+m g) \pm \sqrt{(k L+m g)^2-k^2 L^2}}{k} \\
&=\frac{(k L+m g) \pm \sqrt{2 m g k L+m^2 g^2}}{k}
\end{aligned}
$$
Retain the positive sign.
$$
\therefore y=\frac{(k L+m g)+\sqrt{2 m g k L+m^2 g^2}}{k}
$$
(b) If SHM, at maximum velocity at its lowest point is attained when the body passes, through the "equilibrium, position" $S$ the instantaneous acceleration is zero. That is $m g=k x-\theta$, where $x$ is the extension from $L$.
So the spring or storing force $K x$ is balanced by gravitational force $m g$.
So these two forces will be equal and opposite. $m g=k x \quad$...(i)
Let the maximum velocity of stone be $v$ at bottom of journey. Then,
$$
\frac{1}{2} m v^2+\frac{1}{2} k x^2=m g(L+x)
$$
(By law of conservation of energy, KE of stone + PE gain by string $=$ PE last by stone from $P$ to $Q$ )
$$
\frac{1}{2} m v^2=m g(L+x)-\frac{1}{2} k x^2
$$
Now, $m g=k x \quad$ from (i)
$$
\begin{aligned}
&\Rightarrow x=\frac{m g}{k} \\
&\therefore \frac{1}{2} m v^2=m g\left(L+\frac{m g}{k}\right)-\frac{1}{2} k \frac{m^2 g^2}{k^2} \\
&\quad=m g L+\frac{m^2 g^2}{k}-\frac{1}{2} \frac{m^2 g^2}{k} \\
&\frac{1}{2} m v^2=m g L+\frac{1}{2} \frac{m^2 g^2}{k} \\
&m v^2=2 m g L+\frac{m^2 g^2}{k} \\
&\Rightarrow m v^2=m\left[2 g L+\frac{m g^2}{k}\right] \\
&\therefore v^2=2 g L+m g^2 / k \Rightarrow v=\left(2 g L+m g^2 / k\right)^{1 / 2}
\end{aligned}
$$
(c) If stone is at the lowest point, i.e., at instantaneous distance $Y$ from $P$, then equation of motion of the stone is
$$
\begin{aligned}
&F=m g(\downarrow)-k(y-L) \uparrow \\
&\therefore \frac{m d^2 y}{d t^2}=m g-k(y-L) \\
&\Rightarrow \frac{d^2 y}{d t^2}+\frac{k}{m}\left[\left(y-L-\frac{g}{F}\right]=0\right.
\end{aligned}
$$
Make a transformation of variables,
$$
\begin{aligned}
&z=\frac{k}{m}\left[(y-L)-\frac{g m}{k}\right] \\
&\therefore \frac{d^2 z}{d t^2}+\frac{k}{m} z=0
\end{aligned}
$$
It is a differential equation of second order which represents SHM.
Comparing with equation,
$$
\frac{d^2 z}{d t^2}+\omega^2 z=0
$$
where $\omega$ is angular frequency of harmonic motion, so
$$
\omega=\sqrt{\frac{k}{m}}
$$
The solution of above differential equation will be of type $z=A \cos (\omega t+\phi)$;
where $\omega=\sqrt{\frac{k}{m}}$ and $\phi$ is phase difference.
$$
z=\left(L+\frac{m g}{k}\right)+A^{\prime} \cos (\omega t+\phi)
$$
So, the stone will perform SHM with angular frequency $\omega=\sqrt{k / m}$ about a point $y=0$,
$$
\begin{aligned}
&\left|z_0\right|=\left|-\left(L+\frac{m g}{k}\right)\right| \quad \text { [from(i)] } \\
&z_0=L+\frac{m g}{k}
\end{aligned}
$$