[TIKZ]
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:C] (C) at (12, 0);
\coordinate[label=above:A] (A) at (3,10);
\coordinate (F) at (3,0);
\coordinate (E) at (0.991,3.303);
\coordinate[label=above: P] (P) at (3,2.7);
\coordinate[label=above: D] (D) at (15,10);
\draw (A) -- (B)-- (C)-- (D) -- (A);
\draw (C) -- (E);
\draw (A) -- (F);
\draw (P) -- (D);
\draw (F) rectangle +(-0.2, 0.2);
\draw[thick,dashed] (9,6.35) circle (7.023cm);
\begin{scope}[shift={(1,0)}]
\node[draw,rectangle,rotate=71] at (0.16,3.395){};
\end{scope}
[/TIKZ]
First note that $P$ is the orthocenter of $\triangle ABC$. Furthermore, note that from the perpendicularity $DA\perp AP$ and $DC\perp CP$, so quadrilateral $DAPC$ is cyclic. Furthermore, $DP$ is a diameter of circle $(DAPC)$. This is the circumcircle of $\triangle DAC$, which is congruent to $BCA$. As a result, if $R$ is the circumradius of $\triangle ABC$, then $PD=2R$.
Now I claim that $PB=2R\cos B$. To prove this, reflect $P$ across $AB$ to point $P'$. It is well-known that $P'$ lies on the circumcircle of $\triangle ABC$, so in particular the circumradii of $\triangle APB$ and $\triangle ACB$ are equal. But then by Law of Sines \[\dfrac{BP}{\sin\angle BAP}=\dfrac{BP}{\cos B}=2R\quad\implies\quad BP = 2R\cos B\]as desired. (An alternate way to see this is through the diagram itself: from right triangle trigonometry on triangles $DAP$ and $DCP$ it is not hard to see that $PA=2R\cos A$ and $PC=2R\cos C$, which by symmetry suggests $PB=2R\cos B$.)
Finally, note that by Law of Sines again we have $AC=2R\sin B$, so \[AC^2+BP^2=(2R\sin B)^2 + (2R\cos B)^2 = (2R)^2(\sin^2 B+\cos^2 B) = PD^2.\]Hence \[AC^2=PD^2-PB^2=821^2-700^2=(821-700)(821+700)=11^2\cdot 39^2\]and so $AC=11\cdot 39=\boxed{429}$.