Given a coin is tossed four times

$\therefore$ Possible outcomes =$2^4=16$

$\therefore$ Sample space can be written as

Sample space =HHHH,Amount =1+1+1+1=4

Sample space =HHHT,Amount =1+1+1-1.50=3-1.50=1.50

Sample space =HHTH,Amount =1+1-1.50+1=1.50

Sample space =HHTT,Amount =1+1-1.50-1.50=-1.00

Sample space =HTHH,Amount =1-1.50+1+1=1.50

Sample space =HTHT,Amount =1-1.50+1-1.50=-1.00

Sample space =HTTH,Amount =1-1.50-1.50+1=-1.00

Sample space =HTTT,Amount =1-1.50-1.50-1.50=-3.50

Sample space =THHH,Amount =-1.50+1+1+1=1.50

Sample space =THHT,Amount =-1.50+1+1-1.50=-1.00

Sample space =THTH,Amount =-1.50+1-1.50+1=-1.00

Sample space =THTT,Amount =-1.50+1-1.50-1.50=-3.50

Sample space =TTHH,Amount =-1.50-1.50+1+1=-1.00

Sample space =TTHT,Amount =-1.50-1.50+1-1.50=-3.50

Sample space =TTTH,Amount =-1.50-1.50-1.50+1=-3.50

Sample space =TTTT,Amount =-1.50-1.50-1.50-1.50=-6

Therefore from the samples we get 5 type of different amounts.

$4,1.50,-1.00,-3.50,-6.00$

4 has occurred 1 time

1.50 has occurred 4 times

-1.00 has occurred 6 times

-3.50 has occurred 4 times

-6.00 has occurred 1 time

$\Rightarrow$ P(winning of Rs.4.00)=$\large\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$\Rightarrow \large\frac{1}{16}$

$\Rightarrow$P(winning of Rs1.50)=$\large\frac{4}{16}=\frac{1}{4}$

$\Rightarrow$P(winning of Rs-1.00)=$\large\frac{6}{16}=\frac{3}{8}$

$\Rightarrow$P(winning of Rs-3.50)=$\large\frac{4}{16}=\frac{1}{4}$

$\Rightarrow$P(winning of Rs-6.00)=$\large\frac{1}{16}$

Hence (A) is the correct answer.