Derive relation between Kc and Kn. (plz answer fast)

Relationship between K_p and K_c
Consider the following reversible reaction:
aA + bB ⇌ cC + dD
The equilibrium constant for the reaction expressed in terms of the concentration (mol / litre) may be expressed as:
K _c = [C] ^c [D] ^d / [A] ^a [B] ^b

If the equilibrium involves gaseous species, then the concentrations may be expressed in terms of partial pressures of the gaseous substance. The equilibrium constant in terms of partial pressures may be given as:
K _p = p^c_C p^d_D / p^a_A p^b_B
Where p_A, p_B, p_C and p_D represents the partial pressures of the substance A, B, C and D respectively. If gases are assumed to be ideal, then according to ideal gas equation:
pV = nRT
p = nRT / V
Where p ———-> pressure in Pa
n ——————–> amount of gas in mol
V ——————–> Volume in m³
T ———————> temperature in Kelvin
n/V = concentration, C
or
p = CRT or [gas] RT

If C is in mol dm^-3 and p is in bar, then R = 0.0831 bar dm³ mol^-1 K^-1

Therefore, at constant temperature, pressure of the gas P is proportional to its concentration C, i.e.

Let us suppose a general reaction:
aA + bB↔ cC + dD
The equilibrium constant will be given as:
K_p = (p_C) ^c (p_D) ^d /
(p_A) ^a (p_B) ^b ……. (1)
Now, p = CRT
Hence,
p^A = [A] RT
where [A] is the molar concentration of A
Similarly,
p^B = [B] RT
p^C = [C] RT
p^D = [D] RT
where [B], [C] and [D] are the molar concentration of B, C and D respectively
Substituting these values in expression for K_p i.e. in equation (1)

K_p = [([C] RT) ^c ([D] RT) ^d]/[([A] RT) ^a ([B] RT) ^b]

= [C] ^c [D] ^d (RT) ^c+d/[A] ^a [B] ^b (RT) ^a+b

= [C] ^c [D] ^d (RT) ^{c+d – a+b}/[A] ^a [B] ^b

= K_c (RT) ^{c+d – a+b}
= K_c (RT) ^∆n
Where ∆n = (c + d) – (a + b) i.e. number of moles of gaseous products – number of moles of gaseous reactants in the balanced chemical reaction.

Hence relation between K_p and K_c is given as:
K_p = K_c (RT) ^∆n