The equation used is:
Kp = Kc(RT)^change of n
Kp = Kc(RT)^change of n

In these problems "n" refers to the moles of gases present. Since the reactant is a solid and both products are gases, change in n = 2  0 = 2

Because the reactant is a solid (n = 0) and there are 2 moles of gas in the product (n = 2).
∴ ∆n = 2 – 0 = 2
The equation you're cited is easy to derive:
Kp = P(NH₃) • P(HBr) = {n(NH₃)RT/V} • {n(HBr)RT/V} = {n(NH₃)/V}•{n(HBr)/V}•(RT)²
= [NH₃][HBr]•(RT)² = Kc •(RT)²
∴ ∆n = 2 – 0 = 2
The equation you're cited is easy to derive:
Kp = P(NH₃) • P(HBr) = {n(NH₃)RT/V} • {n(HBr)RT/V} = {n(NH₃)/V}•{n(HBr)/V}•(RT)²
= [NH₃][HBr]•(RT)² = Kc •(RT)²