jmc

There are a total of $6!$ ways to arrange the kids. Abby and Bridget can sit in 3 ways if they are adjacent in the same column: \begin{eqnarray} \text{A}&\quad\text{X}\quad&\text{X} \\ \text{B}&\quad\text{X}\quad&\text{X} \end{eqnarray} \begin{eqnarray} \text{X}&\quad\text{A}\quad&\text{X} \\ \text{X}&\quad\text{B}\quad&\text{X} \end{eqnarray} \begin{eqnarray} \text{X}&\quad\text{X}\quad&\text{A} \\ \text{X}&\quad\text{X}\quad&\text{B} \end{eqnarray} For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in $4!$ ways which results in a total of $3\times 2\times 4!$ ways to arrange them. By the same logic, there are 4 ways for Abby and Bridget to be placed if they are adjacent in the same row: they can swap seats, and the other $4$ people can be arranged in $4!$ ways for a total of $4\times 2\times 4!$ ways to arrange them. We sum the 2 possibilities up to get $\frac{(3\cdot 2)\cdot 4!+(4\cdot 2)\cdot 4!}{6!}=\frac{14\cdot 4!}{6!}=\boxed{\frac{7}{15}}$ or $\text{(C)}$.