From the given data,

$C _{ V }=\frac{3 R ( a + aRT )}{2}$

So,

$C _{ p }= C _{ V }+ R =\frac{3 R ( a + aRT )}{2}+ R =\frac{3 R ( a + aRT )+2 R }{2}$

So,

$\gamma=\frac{C_{ P }}{C_{ V }}-1=\frac{\frac{3 R(a+a R T)+2 R}{2}}{\frac{3 R(a+a R T)}{2}}-1=\frac{2}{3(1+a R T)}=\frac{2}{3}(1+a R T)^{-1}$

Now, by putting this value of $\gamma-1$ in the adiabatic expression formula, we get $TV ^{\gamma-1}=$ constant $\Rightarrow TV ^{3 / 2} e ^{ aRT }=$ constant (we get this by binomial expansion).

Hence proved.