The confusion of entropy change The entropy is defined as randomness of a substance or a system.
The randomness increases as we go from solids to liquids to gases. Similarly from hard diamond to soft graphite. Since graphite is soft and slippery, the orderness would be less than diamond.

But acc. to the formula for entropy change during phase transition,
$\Delta S=Q\ \text{(at const. pressure)}/T$
$\Delta S=\Delta H/T\ \text{(at const. pressure)}$

But $\Delta H$ of reaction/process
C graphite --> C diamond is positive
Meaning $\Delta S$ is positive. But it should be negative due to less entropy of diamond.

Where is the problem?
The standard free energy of formation of graphite is $0\ \mathrm{kJ/mol}$, and the standard free energy of formation of diamond is $2.9\ \mathrm{kJ/mol}$.  So, at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{bar}$, graphite is favored thermodynamically over diamond.  Let's see what pressure would be required for $\Delta G$ to be zero at $25\ \mathrm{^\circ C}$.  Since the temperature is constant, the change in $G$ is described by $\mathrm dG=v\,\mathrm dP$, where $v$ is the molar volume.  The density of graphite is $2.26\ \mathrm{g/cm^3}$ and the density of diamond is $3.51\ \mathrm{g/cm^3}$.  So the molar volumes are $5.31\times 10^{-6}\ \mathrm{m^3/mol}$ for graphite and $3.42\times 10^{-6}\ \mathrm{m^3/mol}$ for diamond.  So, at pressure $P$ (in Pa), the Gibbs free energy for graphite and diamond are:
$$G_\text{Graphite}=0+5.31\times 10^{-6}P$$
$$G_\text{Diamond}=2900+3.42\times 10^{-6}P$$
So, $\Delta G$ at pressure $P$ is $$\Delta G_\mathrm{G\rightarrow D}=2900-1.89\times 10^{-6}P$$So the change in free energy is zero at $P=1.53\times 10^{9}\ \mathrm{Pa}=15\,300\ \mathrm{bar}$

Diamonds are formed naturally at even higher pressures and higher temperatures.  But this is the approximate order of magnitude.

So at $25\ \mathrm{^\circ C}$, graphite is highly thermodynamically favored over diamond.  The actual presence of diamonds at $25\ \mathrm{^\circ C}$ must therefore be a matter of kinetics, rather than thermodynamics.