jmc

The areas bounded by the unit square and alternately bounded by the lines through $\left(\frac{5}{8},\frac{3}{8}\right)$ that are vertical or have a slope of $1/2$ show where $P$ can be placed to satisfy the condition. One of the areas is a trapezoid with bases $1/16$ and $3/8$ and height $5/8$. The other area is a trapezoid with bases $7/16$ and $5/8$ and height $3/8$. Then,$$\frac{\frac{1}{16}+\frac{3}{8}}{2}\cdot \frac{5}{8}+\frac{\frac{7}{16}+\frac{5}{8}}{2}\cdot \frac{3}{8}=\frac{86}{256}=\frac{43}{128}⟹43+128=\boxed{171}$$