W = 87; I = 73; S = 83; K = 75
This yields the following system of equations:
$\begin{cases}
87a+b \equiv 64066 \pmod {256256}\\
73a+b \equiv 158368 \pmod {256256}\\
83a+b \equiv 92525 \pmod {256256}\\
75a+b \equiv 143358 \pmod {256256}
\end{cases}$
The following proposition is useful.
Proposition If $x \equiv y \pmod {n}$ then $x \equiv y \pmod {n/d}$ for any divisor $d$ of $n$.
You can therefore solve the above system modulo the factors of $256256 = 2^8 \cdot 7 \cdot 11 \cdot 13$.
For example, modulo $7$, we obtain:
$\begin{cases}
3a+b \equiv 2\pmod {7}\\
3a+b \equiv 0 \pmod {7}\\
6a+b \equiv 6 \pmod {7}\\
5a+b \equiv 5 \pmod {7}
\end{cases}$
The two first equations (modulo $7$) are impossible. This means that $a$ and $b$ cannot be recovered modulo $7$.
Let us now look modulo $11$:
$\begin{cases}
10a+b \equiv 2\pmod {11}\\
7a+b \equiv 1 \pmod {11}\\
6a+b \equiv 4 \pmod {11}\\
9a+b \equiv 6 \pmod {11}
\end{cases}$
The two first equations yield $a\equiv 4 \pmod {11}$ and $b\equiv 6 \pmod {11}$. However, these solutions are incompatible with the two last equations.
I suspect that there are some errors in the problem. Can you check the values of the ciphertext. Or I made mistakes in the calculation...;)
