Getting the pressure of a system by series expansion
Since this is clearly an exercise, I don't think it's appropriate to give the full solution, but I see no harm in pointing you in the right direction.

You are aiming to convert an expansion in $z$ into an expansion in the number density $\rho=N/A$. These are often called virial expansions. Your first equation can be written
$$
\rho = q\left[z+\frac{z^2}{4}\right] \qquad\text{where}\qquad
q = \frac{2\pi}{(h\beta c)^2} .
$$
I have defined $q$ just to cut down on needless clutter.
The brute force approach to this would be to solve this equation, to give $z$ in terms of $\rho$,
and then substitute this expression into your series formula for $P/K_BT$ as a function of $z$. You can do this; it's a bit fiddly, since it is a quadratic equation in $z$, but it could be done.

For your purposes, a slightly more straightforward approach is to assume that you will get a result of the form
$$
z = c_1 \rho + c_2 \rho^2 + \ldots
$$
where you need to determine the coefficients $c_1$, $c_2$ etc.
Hopefully you can see why there is no need for a $\rho$-independent term $c_0$, when you look at the equation for $\rho$.
Also, it should be obvious that you don't need to consider more terms like $c_3$ etc, if you are only interested in the final result having the first couple of powers of $\rho$.

So you determine the coefficients $c_1$, $c_2$ by substituting that expression for $z$ into the right hand side of the equation for $\rho$,
and equating the coefficients of corresponding powers of $\rho$ on each side.
The term linear in $\rho$ will give the coefficient $c_1$ immediately:
$$
[\rho^1]:\qquad 1 = q \, c_1 .
$$
The quadratic term in $\rho$ will give an equation for $c_2$, which
also involves $c_1$, 
$$
[\rho^2]:\qquad 0 = \text{I'll leave you to work this out.}
$$
Since you just determined $c_1$, this will be easy to solve to get $c_2$.

So now you have $z\approx c_1 \rho + c_2 \rho^2$ and you can insert this expression into the given series expansion for $P/K_BT$,
only keeping the terms in $\rho$ and $\rho^2$.
If you do all this carefully, you will get the answer provided.