Probability of $D_k(a*k+b) =p$ for a PRP
Let's look at some simpler relations first. I'll be sloppy with the probabilities, I'm only trying to gauge the order of magnitude.

The probability that $E_{k_1}(p) = k_1$ is $2^{-128}$ for a given $p$, but must be true for some $p$ with every AES-128 key $k_1$.


The probability that $E_{k_2}(p) = k_2$ for the same $p$ is $2^{-128}$ for a given $k_2$. 
The probability that there is some $k_2 \not = k_1$ for which $E_{k_2}(p) = k_2$ is about $2^{128}-1$ times as much, i.e. quite likely.
The probability that there are three different $k_i$ for which it is true is again close to $2^{-128}$.


Now that was with a constant $p$ that we "found" with some key $k_1$. Take another key and there's another $p'$ for which a similar equality holds, and $p'$ is likely different. For that you can again look at all other keys, probably finding at most one other key that fulfills it, but finding three with probability around $2^{-128}$.

Go through all $p'$, of which there are a small factor fewer than $2^{128}$, and you are quite likely to find one for which three keys match $E_{k_i}(p) = k_i$. Not 100% likely, but maybe 10% or whatever.

Now you can make the equation more complex. If you XOR a random constant $x_1$ on the right side, you can approximate it as an independent equation. There are $2^{128}$ possible values of $x_1$ so now you are overwhelmingly likely to find one triplet of keys for which it holds. Might even find four, but probably not.

Your form is more complex so there's more freedom to choose the constants and $N$ is probably larger. However, my math was sloppy enough already that I won't guess at the value.