Chemical Equilibrium - finding equilibrium constant given degree of dissociation
It looks like alpha $\alpha$ is the degree of dissociation. The way it is explained in your image is a little odd, so let me see if I can do a better job. What seems like it might be problematic for you is that the explanation you have been given skips some algebra along the way. Below, I am showing all of the algebra up until your points of confusion.

For your reaction: $\ce{N2O4 &lt;=&gt; 2NO2}$

In the first part of your question, we need to determine the equilibrium constant 

$$K_p=\frac{P^2_{\ce{NO2}}}{P_{\ce{N2O4}}}$$

We know the total pressure, but we do not know the partial pressures of each species. Fortunately, thanks to Raoult's Law, we can substitute in mole fractions since $P_i =P \chi_i $.

$$K_p=\frac{P^2 \chi^2_{\ce{NO2}}}{P\chi_{\ce{N2O4}}}=P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

We aren't given any number of moles, but let's say that initially there was some number of moles of $\ce{N2O4}$ before any dissociation occurred. We'll call that number $n_0$. We can write expressions for the number of moles of $\ce{N2O4}$ and $\ce{NO2}$ after dissociation using $n_0$ and $\alpha$. Since $\alpha\%$ of $\ce{N2O4}$ is dissociated, $(1-\alpha\%)$ remains undissociated:

$$n_{\ce{N2O4}}=n_0 (1-\alpha)$$

The moles of $\ce{NO2}$ are $n_0$ times $2\alpha$, since $\alpha\%$ of $\ce{N2O4}$ was converted into $\ce{NO2}$ and there are two $\ce{NO2}$ molecules for every $\ce{N2O4}$ molecule:

$$n_{\ce{NO2}} = 2n_0 \alpha$$

The total moles (need this for mole fraction) is 

$$\sum{n}=n_{\ce{N2O4}} + n_{\ce{NO2}} = n_0 (1-\alpha) + 2n_0\alpha = n_0 (1-\alpha +2 \alpha) = n_0 (1+\alpha)$$

When we calculate the mole fractions of each species, the $n_0$ term cancels out, which is good since we do not know what its value is:

$$\chi_{\ce{N2O4}}=\frac{n_{\ce{N2O4}}}{\sum{n}}=\frac{n_0 (1-\alpha)}{n_0 (1+\alpha)}=\frac{1-\alpha}{1+\alpha}$$
$$\chi_{\ce{NO2}}=\frac{n_{\ce{NO2}}}{\sum{n}}=\frac{2n_0\alpha}{n_0 (1+\alpha)}=\frac{2\alpha}{1+\alpha}$$


  Now you know where $1+\alpha$ came from. These values are plugged into the law of mass action for $K_p$ to determine the equilibrium constant. 


In the second part of the question, you use the value of the equilibrium constant and the new pressure to determine a new value of $\alpha$. It all starts with the law of mass action and we manipulate some algebra from there. First the law of mass action using mole fractions:

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

Next, substitute in the expressions for mole fraction in terms of $\alpha$.

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}=P\frac{\frac{(2\alpha)^2}{(1+\alpha)^2}}{\frac{1-\alpha}{1+\alpha}}$$

The term $1+\alpha$ appears in both the numerator and the denominator and can be canceled. 

$$\require{cancel} 
K_p = P\frac{\frac{(2\alpha)^2}{\cancel{(1+\alpha)}(1+\alpha)}}{\frac{1-\alpha}{\cancel{(1+\alpha)}}}=P\frac{(2\alpha)^2}{(1-\alpha)1+\alpha}=P\frac{(2\alpha)^2}{1-\alpha^2}$$

Now we move $P$ over to the left side of the equation to separate the values we know ($K_p$ and $P$) from those we do not know ($\alpha$).

$$\frac{K_p}{P}=\frac{(2\alpha^2)}{1-\alpha^2}$$


  Now you have something like your second equation that you are confused about. Perhaps there is a mistake in the original source?