8. There are two decks of playing cards, each arranged in the following order: the 1st card is the Big Joker, the 2nd card is the Small Joker, followed by the four suits of Spades, Hearts, Diamonds, and Clubs, with each suit arranged in the order of $\mathrm{A}, 2,3, \cdots, \mathrm{J}, \mathrm{Q}, \mathrm{K}$. Someone stacks the two decks of cards in the above order on top of each other, and then from top to bottom, discards the 1st card, places the 2nd card at the bottom, discards the 3rd card, places the 4th card at the bottom... and so on, until only one card remains. What is the remaining card?
8. The 6 of Diamonds in the second deck.

According to the problem, if the number of cards is $2, 2^{2}, 2^{3}, \cdots$, $2^{n}$, then, according to the specified operation method, the last remaining card will be the last card of these cards. For example, if you have only 64 cards, according to the specified operation method, the last remaining card will be the 64th card.

Now, you have 108 cards, which is $108-64=44$ (cards) more. If you follow the specified operation method and discard 44 cards first, then you will have exactly 64 cards left, and the 88th card in the original order will be at the bottom of the remaining cards. This way, if you continue the discard and keep operations, the last remaining card will be the 88th card in the original order. According to the order of the suits in two decks of cards, $88-54-2-26=6$, so the last remaining card is the 6 of Diamonds in the second deck.