As $ t $ varies over all real numbers, \[\begin{pmatrix} 4 \\ -1 \end{pmatrix} + t \begin{pmatrix} 5 \\ 2 \end{pmatrix}\]takes on all points on a line with direction $ \begin{pmatrix} 5 \\ 2 \end{pmatrix} $, and as $ s $ varies over all real numbers, \[\begin{pmatrix} 8 \\ k \end{pmatrix} + s \begin{pmatrix} -15 \\ -6 \end{pmatrix}\]takes on all points on a line with direction $ \begin{pmatrix} -15 \\ -6 \end{pmatrix} $. Since the given equation has infinitely many solutions in $ t $ and $ s $, geometrically, it means that the two lines intersect at infinitely many points. This is possible only if the lines coincide. Note that this can occur, because the direction vector $ \begin{pmatrix} 5 \\ 2 \end{pmatrix} $ of the first line is a scalar multiple of the direction vector $ \begin{pmatrix} -15 \\ -6 \end{pmatrix} $ of the second line. So to find $ k $, we can set $ s $ to be any particular value we like. For convenience, we set $ s = 0 $. Then \[\begin{pmatrix} 4 \\ -1 \end{pmatrix} + t \begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ k \end{pmatrix}.\]The left-hand side becomes \[\begin{pmatrix} 5t + 4 \\ 2t - 1 \end{pmatrix} = \begin{pmatrix} 8 \\ k \end{pmatrix}.\]Then $ 5t + 4 = 8 $ and $ k = 2t - 1 $. Solving for $ t $, we find $ t = \frac{4}{5} $, so $ k = \boxed{\frac{3}{5}} $.