$\underline{\textbf{Answer:}\Rightarrow}\;\;{\textbf{(d)}}$
$\underline{\textbf{Explanation:}\Rightarrow}$
Let's assume the addresses in terms of $\mathrm{x100, x200,\dots}$
Now,
$\mathrm{var1} = \underset{\mathrm x100}{\bbox[lightblue,5px,border: 2px solid red]{5}}\;\;$, $\mathrm{var2} = \underset{\mathrm x200}{\bbox[lightblue,5px,border: 2px solid red]{10}}$
$\mathrm{ptr1} = \underset{\mathrm x300}{\bbox[lightblue,5px,border: 2px solid red]{\mathrm x100}}\;\;$, $\mathrm{ptr2} = \underset{\mathrm x400}{\bbox[lightblue,5px,border: 2px solid red]{\mathrm x200}}\;\;$
Now, on calling the function $\underline{\mathbf{rer(\&ptr1, \;\&ptr2)}}$
$\require{cancel}{\mathrm{ptr2} = \underset{\mathrm x500}{\bbox[lightblue,5px,border: 2px solid red]{\cancel{\mathrm x300}_{\mathrm x400}}}\;\;,\mathrm{ptr1} = \underset{\mathrm x600}{\bbox[lightblue,5px,border: 2px solid red]{\cancel{\mathrm x400}_{\mathrm x300}}}\;\;}$
$\mathrm{ptr1} = \underset{\mathrm x700}{\bbox[lightblue,5px,border: 2px solid red]{\mathrm x300}}$
Now,
$\mathrm{**ptr1 \color{red}{*}= ** ptr2}$ means it will assign the product of the values of double pointer $\mathbf{ptr2}$ and $\mathbf{ptr1}$ to the double pointer $\mathbf{ptr1}$
$\therefore \;\;\require{cancel} \mathrm{var1} = \underset{\mathrm x100}{\bbox[lightblue,5px,border: 2px solid red]{\cancel{5}_{50}}}\;\;$, $\mathrm{var2} = \underset{\mathrm x200}{\bbox[lightblue,5px,border: 2px solid red]{10}}$
$\therefore $ After executing statement $\mathbf{**ptr2 \;+= \; **ptr1}$ which can be again simplified as
$\mathrm{**ptr2 = **ptr2 + ** ptr1\; = \; 50 + 10 = \mathbf{60}}$
So, after executing the below statement, output will be:
$\color{magenta}{\textbf{printf("%d%d",var2,var1);}}$
$\mathbf{Output = 60, 50}$
$\therefore \mathbf{(d)}$ is the correct option.