Relation Between Kp and Kc

For the reversible reaction $A(g) \rightleftharpoons 2B(g)$: Equilibrium constant $K_C = \frac{k_f}{k_b} = \frac{1}{2500} = 4.0 \times 10^{-4}$. Change in gaseous moles: $\Delta n_g = 2 - 1 = 1$. The relationship between $K_P$ and $K_C$ is $K_P = K_C (RT)^{\Delta n_g}$. At $T = 1000\text{ K}$, $RT = 0.0831 \times 1000 = 83.1\text{ L atm mol}^{-1}$. Therefore, $K_P = (4.0 \times 10^{-4}) \times 83.1 = 0.03324 \approx 0.033$.

$K_p=K_c(RT)^{\Delta n_g}$ with $\Delta n_g=2-1=1$. $3=K_c(0.083\times1000) \Rightarrow K_c=\frac{3}{83}=3.6\times10^{-2}$.

$K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants. $K_p = K_c$ only when $\Delta n_g = 0$.
$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$: $\Delta n_g = 2 - 2 = 0$
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$: $\Delta n_g = 2 - 2 = 0$
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$: $\Delta n_g = 2 - 2 = 0$
$2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$: carbon is a solid and is not counted, so $\Delta n_g = 2 - 1 = 1$ and $K_p = K_c(RT)$
Hence $K_c$ and $K_p$ are not equal for $2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)$.
Note: the worked line in the source PDF treats carbon as a gas and gets $\Delta n_g = 0$, which contradicts its own answer; the equation as printed in the question, with C(s), is used here.