Here,

Lets take the sequence of the unit digit that repeat itself:: Plz see the attach diag

The last digit of power of 2 repeat in a cycle of numbers – 2, 4, 8, 6

The last digit of power of 3 repeat in a cycle of numbers – 3, 9, 7, 1

The last digit of power of 4 repeat in a cycle of numbers – 4, 6

The last digit of power of 7 repeat in a cycle of numbers – 7 , 9 ,3 ,1

The last digit of power of 8 repeat in a cycle of numbers – 8, 4, 2, 6

The last digit of power of 9 repeat in a cycle of numbers – 9,1

Now to the ques.

$(32^{28})(33^{47})(37^{19})$

$32^{28}$ = last digit is 2 and to the power of 28. Since $2^{power}$ repeats after every 4th , hence we divide the power by 4 i.e $\frac{28}{4}$= 7 and reminder =0 . The last digit of $32^{28}$ = 6

Similarly for $33^{47}$ = here $3^{power}$ repeats after every 4th, so $\frac{47}{4}$= 11 and reminder 3, The last digit will be = 7

And for $37^{19}$ = here $7^{power}$ repeats after every 4th, so $\frac{19}{4}$ = 4 and reminder 3, The last digit will be = 3

Combining the digit = $6*7*3 = 126$ ,

the last digit for the whole equation is = 6