How far?

$From\quad the\quad British\quad Flag\quad Theorem,\\ \\ { PA }^{ 2 }+{ PC }^{ 2 }={ PB }^{ 2 }+{ PD }^{ 2 }\\ \\ PC=\sqrt { { PB }^{ 2 }+{ PD }^{ 2 }-{ PA }^{ 2 } } \\ \\ From\quad the\quad figure,\\ PA=7\quad ,\quad PB=5\quad ,\quad PD=8\\ \\ So,\\ \\ PC=\sqrt { { PB }^{ 2 }+{ PD }^{ 2 }-{ PA }^{ 2 } } =\sqrt { 25+64-49 } \\ =\sqrt { 40 } \approx 6.32\\ \\ \quad \quad \quad \quad \quad \boxed { \therefore PC\approx 6.32 } \\ \\$

I have done it by GeGebra. You down load from GeoGebra.org a free-ware

Point P divides AB horizontally into two pieces $a$ and $b$ , and vertically into two pieces $c$ and $d$ .

Now $|CP|^2 = b^2+d^2=(b^2+c^2)+(a^2+d^2)-(a^2+b^2)=5^2+8^2-7^2=40.$ So $CP = \sqrt{40} \approx \boxed{6.325}.$