The correct order of the size of C, N,P  and S is

C and N are of 2nd period element. Size decreases from C to N.
Therefore, size of N < C.
P and S are of 3rd period elements. But size of p > S [opposite to the expected trend, i.e. size of S > P].
Because P has half-filled stable configuration.
Increase of one electron in S causes repulsion between paired electrons $\left(3s^{2}3p_x^{2}3p_y^{1}3p_z^1\right)$ and thus size of s is slightly increases from P to S.
Moreover, size of 3rd period elements is greater than that of 2nd period.
Hence the correct order of sizes is

$\frac{\mathrm{P}>\mathrm{S}}{3\text{ rd period }}>\frac{\mathrm{C}>\mathrm{N}}{2\text{ nd period }}$