Maps.

Label the townships as shown. There are four choices for colors for township $A$ , and three choices for colors for township $B.$

If townships $C$ and $D$ have the same color, then there are two choices for the color of $C$ and $D$ , and there are two choices for colors for each of the remaining townships $E, F, G \text{ and } H$ .

If townships $C \text{ and } D$ have different colors, then there are two choices for the color of $C$ and one choice for a color $D$ .

Since $B,C \text{ and } D$ will have different colors, there is only one choice of color for township $E$ . Then there are two choices for colors for each of the remaining townships $F, G, \text{ and } H$ .

Thus, the total number of ways to color the $8$ townships is: $(4\times3\times2\times2\times2\times2\times2) + (4\times3\times2\times1\times2\times2\times2) = \boxed{576}.$