解：
         设P点的坐标为$(x_0, y_0)$，则：
         $P到l_1的距离d_1= \begin{vmatrix} x_0 + 2 \end{vmatrix} $ 因为$x_0=\frac {y_0^2}8 \geq 0$ 所以$d_1=x_0+2$ $$P到l_2的距离d_2= \begin{vmatrix} \frac {12x_0 - 5y_0 + 15}{\sqrt {12^2 + (-5)^2}} \end{vmatrix} = \begin{vmatrix} \frac {12x_0 - 5y_0 + 15}{13} \end{vmatrix} $$ 设两者的距离之和为$d=d_1+d_2$，则 \begin{align} d=& x_0+2+ \begin{vmatrix} \frac {12x_0-5y_0+15}{13} \end{vmatrix}\\ =& \frac {y_0^2}8 +2+ \begin{vmatrix} \frac {12 \times \frac {y_0^2}8 -5y_0+15}{13} \end{vmatrix}\\ =& \frac {y_0^2}8 +2+ \begin{vmatrix} \frac {3 \times \frac {y_0^2}2 -5y_0+15}{13} \end{vmatrix}\\ \end{align} 因为 \begin{align} & \frac32y_0^2-5y_0+15\\ =& \frac 32 \left( y_0^2 - \frac {10}3 y_0 + \frac {25}9 \right)^2 - \frac {25}6 + 15\\ =& \frac 32 \left( y_0 - \frac 53 \right)^2 + \frac {65}6 > 0 \end{align} 所以 \begin{align} d=& \frac {y_0^2}8+2+\frac{3\times \frac {y_0^2}2-5y_0+15}{13}\\ =& \left( \frac 18 + \frac 3{26} \right)y_0^2- \frac 5{13}y_0+\frac {41}{13}\\ =& \frac {25}{104}\left( y_0 - \frac 45 \right)^2+3 \end{align} 从距离的表达式中可以看出，当 $$ y_0=\frac 45 $$ 时，距离最小，为$d=3$
         所以，P点到这两条直线的距离之和的最小值为3