Firstly, it's easy to start off with just the Dirac delta potential and see what that does. Wiki has a nice solution for the Delta fuction potential, and I am lifting off parts of it here.
Consider a potential $V(x) = a\delta (x)$ and consider a scattering like configuration, where a plane wave $e^{ikx}$ is incident from the left.
$$
\psi(x)=\begin{cases}e^{ikx}+re^{-ikx} & x<0 \\ te^{ikx} & x> 0\end{cases}
$$
By matching the boundary conditions, like on the wiki page, you get
$$
t = 1+r\\
(1-\alpha)t = 1-r
$$
where
$$
\alpha = \frac{ 2ma}{ik\hbar^2}
$$
characterizes the effect of the delta potential. Solving for $r$ and $t$,
$$
t = \frac{1}{1-\alpha/2}\\
r=-\frac{\alpha/2}{1-\alpha/2}
$$
Now, it is easy to see that for high incident $k$, the only effect of the dirac delta potential is to write a phase discontinuity on the wavefuction. This is because, as $k$ increases, the transmission $|t|^2=1/(1+|\alpha|^2/4)$ approaches 1, but the transmitted wavefunction gets an extra phase given by
$$
\text{Arg}(t) = -\tan^{-1}(|\alpha|/2)
$$
Getting back to the problem at hand, for a particle in a box (without the delta function), the allowed $k$ vectors are given by forcing the wavefunction to be zero at the walls at $x=-d$ and $x=d$, which gives us the condition
$$
k_n=\frac{\pi n}{2d}
$$
If now, we add a delta potential, then for high values of $n$ (or $k$), all the delta function will do is introduce a phase discontinuity at the origin, and consequently what you should expect is that the boundary condition is matched not for $k_n$, but something slightly off $k_n+\delta k_n$, where $\delta k_n$ is a small correction due to the delta function potential. For high values of $n$, this correction would drop, as the phase discontinuity decreases, and for classical like states (very large $n$) you expect to recover 1D box states, as mentioned by John Rennie.