Answer
2 will win.
Observations:
Player 1 cannot put out any pairs first, as then Player 2 will win by beating it with 9 pair, and then using their final card in the next turn.
As such, P1's goal would be to bait out one of P2's 9s. Beat it with King, and remove the remainder of their hand in pairs. Their pairs will always win now, so after that they will have a chance to put out one last card. As such, If P1 only has at most one non-pair card at this moment, P1 wins.
Conversely, P2's goal would be to make sure that after countering one of your 9s with K, P1 has more than one non-pair cards in their hand.
As such
P2 would win, because there is no possible way for P1 to remove the cards in their hand one-by-one without giving an opportunity for P2 to achieve their win condition. P2 is always able to use both their 7 and 9 to force P1 to lose their winning hand.
P2 should wait until P1 uses their first 4 or 6.
At this moment, if P1 still has an even amount of 8, P2 should use 7.
- If P1 doesn't do anything, you use your 9s as a pair next turn and win.
- If P1 counters with 8, their hand will become suboptimal [e.g. 4, 6, 6, 8, K]. You will win.
- If P1 counters with K, your two 9s will become the highest cards in the table and you will win.
On the other hand, if P1 only has one 8 remaining, P2 should use 9.
- If P1 doesn't do anything, you use your 7 next turn.
Now no matter what P1 does, their hand [e.g. 4, 6, 6, 8, K] cannot win against your 9.
- If P1 counters with K, your remaining 9 will become the highest card in the table.
Since P1 has two individual cards remaining (4 and 8), they cannot win before you.
TL;DR: As long as P2 is patient and plays their card correctly, they will win. P1 has no way of using up all their cards without P2 intercepting and creating their own victory condition.