Let us consider a spring of spring constant $k$ is attached to mass $m$ at one end and other end is fixed at right support.
When spring is stretched by a force $F$ with the displacement $(x)$, restoring force will be $F=-k x$.
(-ve) sign shows that displacement $x$ is opposite direction.
Potential energy of the stretched spring,
$$
P E=\frac{1}{2} k x^2
$$
The restoring force is directly proportional to the $x$, when particle released it will execute SHM about equilibrium position.
Acceleration will be, $a=\frac{-F}{m}=\frac{-k x}{m}$
At equilibrium position, $x=0 \Rightarrow a=0$; at $x=x_{\max }, a_{\max }=\frac{-k x_{\max }}{m}$ when just released. Hence, acceleration is maximum. So option (a) is verifies.
At equilibrium whole PE will be converted to $\mathrm{KE}$ hence, $\mathrm{KE}$ will be maximum and so that speed of mass will be maximum at $x=0$.