The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} {19}&{47} \ {34}&{28} \end{array}} \right]}$.
To find the transpose of a matrix, you swap the rows and columns. So, the transpose of matrix $A$ is $\left[ {\begin{array}{{20}{c}} {2}&{1} \ {4}&{3} \end{array}} \right]$. The transpose of matrix $B$ is $\left[ {\begin{array}{{20}{c}} {4}&{5} \ {6}&{9} \end{array}} \right]$. Therefore, the transpose of the product of these two matrices is $\left[ {\begin{array}{*{20}{c}} {19}&{47} \ {34}&{28} \end{array}} \right]$.
Here is a step-by-step solution:
- Find the product of matrices $A$ and $B$. This is given by the following equation:
$$AB = \left[ {\begin{array}{{20}{c}} {2}&{4} \ {1}&{3} \end{array}} \right] \left[ {\begin{array}{{20}{c}} {4}&{6} \ {5}&{9} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {24}&{42} \ {23}&{41} \end{array}} \right]$$
- Transpose the product matrix. This is given by the following equation:
$$(AB)^T = \left[ {\begin{array}{{20}{c}} {24}&{23} \ {42}&{41} \end{array}} \right]^T = \left[ {\begin{array}{{20}{c}} {24}&{42} \ {23}&{41} \end{array}} \right]$$
- Swap the rows and columns of the resulting matrix. This gives the following matrix:
$$\left[ {\begin{array}{{20}{c}} {24}&{42} \ {23}&{41} \end{array}} \right]^T = \left[ {\begin{array}{{20}{c}} {24}&{23} \ {42}&{41} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {19}&{47} \ {34}&{28} \end{array}} \right]$$
Therefore, the transpose of the product of matrices $A$ and $B$ is $\boxed{\left[ {\begin{array}{*{20}{c}} {19}&{47} \ {34}&{28} \end{array}} \right]}$.