$2^{1}=2 \ ; \ 2^{2} = 4 \ ; \ 2^{3}=8 \ ;\ 2^{4}=16$
$2^{5}=32 \ ; \ 2^{6} = 64 \ ; \ 2^{7}=128 \ ;\ 2^{8}=256$
From above it is clear that the Cyclicity of 2 is 4.
$3^{1}=3 \ ; \ 3^{2} =9 \ ; \ 3^{3}=27 \ ;\ 3^{4}=81$
$3^{5}=243 \ ; \ 3^{6} =729 \ ; \ 3^{7}=2187 \ ;\ 3^{8}=6561$
From above it is clear that the Cyclicity of 3 is 4.
$4^{1}=4 \ ; \ 4^{2} =16 \ ; \ 4^{3}=64 \ ;\ 4^{4}=256$
From above it is clear that the Cyclicity of 4 is 2.
$5^{1}=5 \ ; \ 5^{2} =25 \ ; \ 5^{3}=125 \ ;\ 5^{4}=625$
From above it is clear that the Cyclicity of 5 is 1.
Hence,
Last digit of $22^{2}=4$
Last digit of $33^{3}=3$
Last digit of $44^{4}=6$
Last digit of $55^{5}=5$
The last digit of $22^{2} \times 3^{33} \times 44^{4} \times 55^{5}=4 \times 3 \times 6 \times 5 = 360$
Hence,The last digit of $22^{2} \times 3^{33} \times 44^{4} \times 55^{5} = 0$
Hence, Option(d)0 is the correct choice.
