Solution:

Answer: $\frac{3}{4}$

We first prove that $ONPM$ is a parallelogram. Note that $APD$ and $BPC$ are both $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Let $M'$ denote the intersection of $MP$ and $BC$. Since $\angle BPM' = \angle MPD = 30^{\circ}$, we have $MP \perp BC$. Since $ON$ is the perpendicular bisector of $BC$, we have $MP \parallel NO$. Similarly, we have $MO \parallel NP$. Thus $ONPM$ is a parallelogram.

It follows that the area of triangle $MON$ is equal to the area of triangle $MPN$, which is equal to $\frac{1}{2} \cdot 1 \cdot 3 \cdot \sin \angle MPN = \frac{1}{2} \cdot 1 \cdot 3 \cdot \sin 150^{\circ} = \frac{3}{4}$.