At 600 K the equilibrium constant of N₂ + 3H₂ ⇌ 2NH₃ is Kp = 4.43 × 10⁻³ bar⁻²…

Step 1. Amounts. N2: $1 - \xi$. H2: $3 - 3\xi$. NH3: $2\xi$. Total: $4 - 2\xi$.

Step 2. The expression. The factors of $4 - 2\xi$ cancel in the quotient, because the powers in the numerator and the denominator differ by 2: $K_p P^2 = \dfrac{(2\xi)^2 (4 - 2\xi)^2}{(1 - \xi)(3 - 3\xi)^3} = \dfrac{16\,\xi^2 (2 - \xi)^2}{27\,(1 - \xi)^4}$.

Step 3. Square root. $\sqrt{K_p}\,P = \dfrac{4\,\xi\,(2 - \xi)}{\sqrt{27}\,(1 - \xi)^2}$. With $\sqrt{4.43 \times 10^{-3}} = 0.06656$, the left side is $0.06656 \times 100 = 6.656$. Put $a = 6.656 \times \dfrac{\sqrt{27}}{4} = 8.646$.

Step 4. Solve. Then $\xi(2 - \xi) = a (1 - \xi)^2$. Because $\xi(2 - \xi) = 1 - (1 - \xi)^2$, we get $1 = (1 + a)(1 - \xi)^2$. So $\xi = 1 - \dfrac{1}{\sqrt{1 + a}} = 1 - \dfrac{1}{\sqrt{9.646}} = 0.678$.

Step 5. Ammonia. $y_{\text{NH}_3} = \dfrac{2\xi}{4 - 2\xi} = \dfrac{1.356}{2.644} = 0.513$.