Note \(4-2\sqrt3=2(2-\sqrt3)\). Divide both sides:
\[p+\frac{\sqrt3}{2}=\frac{3-\sqrt3}{4-2\sqrt3}\]
Rationalise the right side by \(\times\dfrac{4+2\sqrt3}{4+2\sqrt3}\): denominator \(16-12=4\); numerator \((3-\sqrt3)(4+2\sqrt3)=12+2\sqrt3-6=6+2\sqrt3\). So the right side \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).
Note \(4-2\sqrt3=2(2-\sqrt3)\). Divide both sides:
\[p+\frac{\sqrt3}{2}=\frac{3-\sqrt3}{4-2\sqrt3}\]
Rationalise the right side by \(\times\dfrac{4+2\sqrt3}{4+2\sqrt3}\): denominator \(16-12=4\); numerator \((3-\sqrt3)(4+2\sqrt3)=12+2\sqrt3-6=6+2\sqrt3\). So the right side \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).