Observe the figure and find $\angle P$.

$(40^{\circ})$ In $\Delta ABC$ and $\Delta PQR$,
$\frac{AB}{RQ} = \frac{3.8}{7.6} = \frac{1}{2}$,$\frac{BC}{QP} = \frac{6}{12} = \frac{1}{2}$ and $\frac{CA}{PR} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2}$
That is,$\frac{AB}{RQ} = \frac{BC}{QP} = \frac{CA}{PR}$
So,$\Delta ABC \sim \Delta RQP$ ($SSS$ similarity criterion).
Therefore,$\angle C = \angle P$ (Corresponding angles of similar triangles).
In $\Delta ABC$,by the angle sum property:
$\angle C = 180^{\circ} - \angle A - \angle B$
$\angle C = 180^{\circ} - 80^{\circ} - 60^{\circ} = 40^{\circ}$
Since $\angle C = \angle P$,we have $\angle P = 40^{\circ}$.