1. Set up coordinates

Take the two perpendicular plane mirrors as the coordinate axes:

• Mirror $M_1$ along the $y$-axis: $x=0$
• Mirror $M_2$ along the $x$-axis: $y=0$

Since the point source $P$ is at distances:

• $a$ from $M_1$
• $2a$ from $M_2$

its coordinates are $P(a,2a).$

2. Find all images formed by two perpendicular mirrors

For two plane mirrors at right angle, three images are formed:

Image in $M_1$

Reflection across $x=0$ changes $x \to -x$: $I_1(-a,2a).$

Image in $M_2$

Reflection across $y=0$ changes $y \to -y$: $I_2(a,-2a).$

Image in both mirrors

Reflecting in both mirrors changes both coordinates: $I_3(-a,-2a).$

3. Compute distances between the images

We need the shortest distance between the images.

Distance between $I_1$ and $I_2$

$I_1(-a,2a), \quad I_2(a,-2a)$ $d_{12}=\sqrt{(a-(-a))^2+(-2a-2a)^2}$ $=\sqrt{(2a)^2+(-4a)^2}$ $=\sqrt{4a^2+16a^2}$ $=\sqrt{20}a=2\sqrt{5}\,a.$

Using $\sqrt{5}=2.3$, $d_{12}=2(2.3)a=4.6a.$

Distance between $I_1$ and $I_3$

$I_1(-a,2a), \quad I_3(-a,-2a)$ $d_{13}=|2a-(-2a)|=4a.$

Distance between $I_2$ and $I_3$

$I_2(a,-2a), \quad I_3(-a,-2a)$ $d_{23}=|a-(-a)|=2a.$

4. Choose the shortest distance

Among $4.6a, \quad 4a, \quad 2a$ the shortest is clearly $2a.$

5. Match with options

The value $2a$ is not present in the options.

So the correct physical result is: $\boxed{2a}$

Hence, the stored answer $\boxed{B}$ is not consistent with the geometry.

A likely reason is that the question intended some restricted pair of images, but as stated, “the images thus formed” means all images formed, whose minimum separation is $2a$ between $I_2$ and $I_3$.