1. Interpret the geometry

Let the original line through $P(a,0)$ make an acute angle $\alpha$ with the positive $x$-axis. So its slope is $m_1=\tan\alpha.$

It is then rotated clockwise by $\dfrac{\alpha}{2}$ about $P$, so the new inclination becomes $\alpha-\frac{\alpha}{2}=\frac{\alpha}{2}.$ Hence the slope of the new line is $m_2=\tan\frac{\alpha}{2}.$

Given: $m_2=2-\sqrt{3}.$ Therefore, $\tan\frac{\alpha}{2}=2-\sqrt{3}.$

2. Find $\alpha$

We use the standard value: $\tan 15^\circ = 2-\sqrt{3}.$ So, $\frac{\alpha}{2}=15^\circ \implies \alpha=30^\circ.$ Thus, $\tan\alpha=\tan 30^\circ=\frac{1}{\sqrt{3}},$ so $\tan^2\alpha=\frac{1}{3}.$

3. Equation of the new line

The new line passes through $P(a,0)$ and has slope $2-\sqrt{3}$, so its equation is $y=(2-\sqrt{3})(x-a).$ Rewriting, $(2-\sqrt{3})x-y-a(2-\sqrt{3})=0.$

4. Use distance from origin

Distance of line $Ax+By+C=0$ from origin is $\frac{|C|}{\sqrt{A^2+B^2}}.$

Here, $A=2-\sqrt{3},\quad B=-1,\quad C=-a(2-\sqrt{3}).$ Given distance is $\dfrac{1}{\sqrt{2}}$, so $\frac{|a|(2-\sqrt{3})}{\sqrt{(2-\sqrt{3})^2+1}}=\frac{1}{\sqrt{2}}.$

Now simplify: $(2-\sqrt{3})^2=4+3-4\sqrt{3}=7-4\sqrt{3},$ so $(2-\sqrt{3})^2+1=8-4\sqrt{3}=4(2-\sqrt{3}).$ Hence, $\sqrt{(2-\sqrt{3})^2+1}=\sqrt{4(2-\sqrt{3})}=2\sqrt{2-\sqrt{3}}.$

So, $\frac{|a|(2-\sqrt{3})}{2\sqrt{2-\sqrt{3}}}=\frac{1}{\sqrt{2}}.$ Multiply both sides: $|a|=\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{2}(2-\sqrt{3})}.$

A cleaner way is to square directly from $\frac{|a|(2-\sqrt{3})}{\sqrt{(2-\sqrt{3})^2+1}}=\frac{1}{\sqrt{2}}.$ Squaring: $\frac{a^2(2-\sqrt{3})^2}{(2-\sqrt{3})^2+1}=\frac12.$ Thus, $2a^2(2-\sqrt{3})^2=(2-\sqrt{3})^2+1.$ But $(2-\sqrt{3})^2+1=4(2-\sqrt{3}),$ so $2a^2(2-\sqrt{3})^2=4(2-\sqrt{3}).$ Divide by $2(2-\sqrt{3})$: $a^2(2-\sqrt{3})=2.$ Therefore, $a^2=\frac{2}{2-\sqrt{3}}=2(2+\sqrt{3})=4+2\sqrt{3}.$

5. Compute the required expression

We need $3a^2\tan^2\alpha-2\sqrt{3}.$ Since $\tan^2\alpha=\frac13,$ we get $3a^2\tan^2\alpha = 3a^2\cdot \frac13 = a^2.$ Hence the expression becomes $a^2-2\sqrt{3}=(4+2\sqrt{3})-2\sqrt{3}=4.$

6. Final answer

The required value is $\boxed{4}.$ So the correct option is B.