If ‘x’, ‘v’ and ‘a’ denote the displacement, velocity and acceleration of a particle respectively executing SHM of periodic time t, then which one of the following does not change with time?

Key Idea for checking the correctness of an equation dimensional analysis is done using the principle of homogeneity.
The dimensions of given variables of SHM are as
Displacement, $\left[x\right]=\left[M^{0}L{T}^0\right]$
Velocity, $\left[v\right]=\left[M^{0}L{T}^{-1}\right]$
Acceleration, $\left[a\right]=\left[M^{0}L{T}^{-2}\right]$
and time period, $\left[T\right]=\left[M^0{L}^{0}T\right]$
Now, checking each option for these values for option (a)
$\frac{\left[a\right]\left[T\right]}{\left[x\right]}=\frac{\left[M^{0}L{T}^{-2}\right]}{\left[M^{0}L{T}^0\right]}=\left[M^0{L}^0{T}^{-1}\right]$
As it depends on time, so change with it.
For option (b),
$\left[a\right]\left[T\right]+2\pi \left[v\right]=\left[M^{0}L{T}^{-2}\right]\left[M^0{L}^{0}T\right]+\left[M^0{L}^0{T}^{-1}\right]$
= $\left[M^{0}L{T}^{-1}\right]$
It is also dependent on time and hence changes with it.
For option $\left(c\right)$ ,
$\frac{\left[a\right]\left[T\right]}{\left[v\right]}=\frac{\left[M^{0}L{T}^{-2}\right]\left[M^0{L}^{0}T\right]}{\left[M^{0}L{T}^{-1}\right]}=\left[M^0{L}^0{T}^0\right]$
As it is a constant having no dimension, so it does not change with time.
For option (d),
$\left[a\right]\left[t\right]+4{\pi }^2{\left[d\right]}^2=\left[M^{0}L{T}^{-2}\right]\left[M^0{L}^{0}T\right]+{\left[M^{0}L{T}^{-1}\right]}^2$
$=\left[L{T}^{-1}\right]+\left[L^2{T}^{-2}\right]$
As, the terms is dependent on time, so changes with it.
Also, it is dimensionally incorrect.
Hence, option $\left(c\right)$ is correct.