Point inside a rectangle

Using the British Flag Theorem , we have

$(PA)^2+(PC)^2=(PB)^2+(PD)^2$

$7^2+24^2=15^2+(PD)^2$

$49+576=225+(PD)^2$

$(PD)^2=400$

$PD=\sqrt{400}=\boxed{20}$

Let $AB=CD=a$ and $AD=BC=b$ , and the coordinates of $A(0,0)$ and $P(x,y)$ . Then the coordinates of $B(a,0)$ , $C(a,b)$ and $D(0,b)$ . By Pythagorean theorem :

$\begin{cases} AP^2 = x^2 + y^2 = 49 & ...(1) \\ BP^2 = (a-x)^2+y^2 = 225 & ...(2) \\ CP^2 = (a-x)^2 + (b-y)^2 = 576 & ...(3) \\ DP^2 = x^2 + (b-y)^2 & ...(4) \end{cases}$

We note that $(3)-(2)+(1)=(4)$ or $CP^2-BP^2+AP^2=DP^2$ . $\implies AP^2+CP^2=BP^2+DP^2$ or British flag theorem as mentioned by @Guillermo Templado .

Therefore, $DP^2 = 576 - 225 + 40 = 400$ , $\implies DP = \boxed{20}$ .

Use British Flag Theorem $\huge DP = \sqrt{7^2 + 49^2 - 15^2} = \sqrt{400} = 20$