Derive an expression for the degree of dissociation, α, in terms of K_p

For the homogeneous gaseous reaction
2 AB₂ (g) ⇌ 2 AB (g) + B₂ (g)
Derive an expression for the degree of dissociation, α, in terms of K_p and total pressure P assuming that α << 1. Given reaction is: 2 AB₂ (g) ⇌ 2 AB (g) + B₂ (g)
Number of moles at Equilibrium:
AB₂ = 2(1 – α)
AB = 2 α
B₂ = α
Total number of moles at equilibrium = 2(1 – α) + 2α + α = 2 + α
K_p = p² (AB) X p (B₂) / p² (AB₂)
P (AB₂) = 2(1 – α) p / 2 + α
P (AB) = 2α p / 2 + α
P (B₂) = αp / 2 + α

= α³ p / (2 + α) (1 – α) ²
As α << 1, it can be neglected in the denominator and we will get: K_p = α³ p / 2
Hence, α = (2 K_p / p) ^1/3