Let money given to sania by his father $= x$
Spent money at 1st shop = $\frac{x}{2}+1=\frac{x+2}{2}$
Remaining money =$x-\frac{x+2}{2}=\frac{x-2}{2}$
Spent money at 2nd shop = $\frac{x-2}{4}+1=\frac{x+2}{4}$
Remaining money =$\frac{x-2}{2}-\frac{x+2}{4}=\frac{x-6}{4}$
Spent money at 3rd shop =$\frac{x-6}{8}+1=\frac{x+2}{8}$
Remaining money =$\frac{x-6}{4}-\frac{x+2}{8}=\frac{x-14}{8}$
Spent money at 4th shop = $\frac{x-14}{16}+1=\frac{x+2}{16}$
Remaining money =$\frac{x-14}{8}-\frac{x+2}{16}=\frac{x-30}{16}$
Spent money at 5th shop = $\frac{x-30}{32}+1=\frac{x+2}{32}$
Remaining money$ =0$
Spent money = Given money
$\frac{x+2}{2}+ \frac{x+2}{4}+ \frac{x+2}{8}+ \frac{x+2}{16} +\frac{x+2}{32}=x$
$\Rightarrow \frac{16x+32+8x+16+4x+8+2x+4+x+2}{32}=x$
$ \Rightarrow \frac{31x+62}{32}=x$
$ \Rightarrow \boxed{x=62}$
Hence,Option(D) is the correct choice.