Calculating Kp in two different ways The degree of dissociation of $\ce{NH3}$ at $\pu{1 atm}$ is 20 % as follows:
     $$\ce{2NH3 &lt;=&gt; N2 + 3H2} $$


I follow two ways but end up with two different answers.

Here, I assume that initial amount of reactant mole is 2 and that of product is 0. Then I take the amount of ammonia in equilibrium to be $2-2\alpha$, nitrogen to be $\alpha$ and hydrogen to be $3\alpha$. Then, I find $K_\mathrm{p}$ (where $\alpha=0.2$).

Here, as dissociation is 20 %, I assume that amount of ammonia in equilibrium will be 0.8 and that of nitrogen and hydrogen will be 0.2 and 0.6 respectively. Then also, I find $K_\mathrm{p}$, but it differs from the initial one.

Where am I mistaken?
First solution is perfect.

The issue is in the second solution.

If we start with 1 mole of $\ce{NH3}$ then with 20% dissociation we are left with 0.8 moles as 0.2 moles react these 0.2 moles give o.1 mole of $\ce{N2}$ and 0.3 moles of $\ce{H2}$ 

Hence $$K_\mathrm p = \frac{[\ce{H2}]^3[\ce{N2}]}{[\ce{NH3}]^2} = \frac{[\ce{0.3}]^3[\ce{0.1}]}{[\ce{0.8}]^2}$$