We have, $BC=2\sqrt{5}.$
Since, $\mathrm{\Delta }\mathrm{}PBC$ is an equilateral triangle and the point $P$ lies in the interior of $\mathrm{\Delta }\mathrm{}ABC.$

Therefore, $P$ lies on the perpendicular bisector of $BC$ at a distance of $\frac{\sqrt{3}}{2}\left(BC\right)=\sqrt{15}$ from the mid-point of $BC$ such that $P$ and $A$ are on the same side of $BC.$
The equation of $BC$ is $x-2y+4=0$
The coordinates of the mid-point of $BC$ are $D\left(\mathrm{0,2}\right)$
The slope of a line perpendicular to $BC$ is $-2.$

So, if it makes an angle $\theta$ with the $x$ -axis. Then,
$\tan \theta =-2\Rightarrow \sin \theta =\frac{2}{\sqrt{5}}$ and $\cos \theta =-\frac{1}{\sqrt{5}}$
The parametric form of the perpendicular bisectors of $BC$ is
$\frac{x-0}{-\frac{1}{\sqrt{5}}}=\frac{y-2}{\frac{2}{\sqrt{5}}}$
So, the coordinates of $P$ are given by
$\frac{x-0}{-\frac{1}{\sqrt{5}}}=\frac{y-2}{\frac{2}{\sqrt{5}}}=\pm \sqrt{15}$
or, $x=\pm \sqrt{3},y=2\pm 2\sqrt{3}$
Thus, the coordinates of $P$ are $\left(\sqrt{3},2-2\sqrt{3}\right)$ or $\left(-\sqrt{3},2+2\sqrt{3}\right).$
Clearly, $A\left(\mathrm{6,7}\right)$ and $\left(-\sqrt{3},2+2\sqrt{3}\right)$ lie on the same side of $BC.$ Hence, the coordinates of $P$ are $\left(-\sqrt{3},2+2\sqrt{3}\right).$