1. Equation of the incident ray

The ray starts from the origin and makes an angle $30^\circ$ with the positive $x$-axis.

So its slope is $m=\tan 30^\circ=\frac{1}{\sqrt{3}}.$

Hence the equation of the incident ray is $y=\frac{x}{\sqrt{3}}.$

2. Point of incidence with the line $x+y=1$

We intersect $y=\frac{x}{\sqrt{3}}$ with $x+y=1.$

Substitute $y=\dfrac{x}{\sqrt{3}}$ into $x+y=1$: $x+\frac{x}{\sqrt{3}}=1.$

So, $x\left(1+\frac{1}{\sqrt{3}}\right)=1$ $x=\frac{1}{1+1/\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}+1}.$

Then $y=\frac{x}{\sqrt{3}}=\frac{1}{\sqrt{3}+1}.$

Thus the point of incidence is $P\left(\frac{\sqrt{3}}{\sqrt{3}+1},\frac{1}{\sqrt{3}+1}\right).$

3. Slope of the reflected ray

The mirror is the line $x+y=1 \implies y=-x+1,$ which has slope $-1$.

Hence it makes an angle $135^\circ$ with the positive $x$-axis.

The incident ray makes angle $30^\circ$. On reflection about a line with angle $\alpha$, the reflected direction angle $\theta_r$ is $\theta_r=2\alpha-\theta_i.$

Here, $\alpha=135^\circ, \qquad \theta_i=30^\circ.$

Therefore, $\theta_r=2(135^\circ)-30^\circ=240^\circ.$

So the reflected ray has slope $m_r=\tan 240^\circ=\tan 60^\circ=\sqrt{3}.$

Since the ray travels from $P$ downward toward the $x$-axis, this is consistent.

Thus the reflected ray through $P$ is $y-\frac{1}{\sqrt{3}+1}=\sqrt{3}\left(x-\frac{\sqrt{3}}{\sqrt{3}+1}\right).$

4. Find intersection with the $x$-axis

On the $x$-axis, $y=0$. So: $0-\frac{1}{\sqrt{3}+1}=\sqrt{3}\left(x-\frac{\sqrt{3}}{\sqrt{3}+1}\right).$

Thus, $-\frac{1}{\sqrt{3}+1}=\sqrt{3}x-\frac{3}{\sqrt{3}+1}.$

So, $\sqrt{3}x=\frac{2}{\sqrt{3}+1}.$

Hence, $x=\frac{2}{\sqrt{3}(\sqrt{3}+1)}=\frac{2}{3+\sqrt{3}}.$

Therefore the abscissa of $Q$ is $\boxed{\frac{2}{3+\sqrt{3}}}.$

5. Option check

This matches Option D.

6. Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

So they agree.