This problem can be solved with dynamic programming. First, define $b_{i,j}$ to represent the $j^{th}$ board on the $i^{th}$ layer for $(1\le i\le N-1)$, or the $j^{th}$ bucket on the $i^{th}$ layer for $i=N$.

Define $dp[i][j]$ to be minimum cost to reach $b_{i,j}$. Note that the ball can come from either $b_{i-1,j}$ or $b_{i-1,j-1}$. Define the cost $c_{left}$ to be $0$ if $b_{i-1,j-1}$ is R or $1$ if $b_{i-1,j-1}$ is L. Similarly, define the cost $c_{right}$ to be $0$ if $b_{i-1,j}$ is L or $1$ if $b_{i-1,j}$ is R. This extra cost is due to the fact that if the left/right board on the previous layer does not direct the ball to the current board, we will have to flip it so that it does.

Our general transition goes something like this:

$${\displaystyle dp[i][j]=min(dp[i-1][j-1]+c_{left},dp[i-1][j]+c_{right})}$$

Now, to find the maximum $P$, we can simply take $max(t_i-dp[N][i])$ where $(1\le i\le N)$.

However, how do we know exactly what boards we flipped throughout the course of our calculation?

The answer is to store the optimal paths taken in a backtracking array. Say $bt[i][j]$ is $0$ if the ball came from $b_{i-1,j-1}$, or $1$ if the ball came from $b_{i-1,j}$ on some optimal path to $b_{i,j}$. This can be calculated while we fill the $dp$ matrix. If our maximum $P$ happens when the ball lands in the bucket at index $idx$, we can start our backtracking from $bt[N][idx]$ and move upwards to $b_{i-1,j}$ or $b_{i-1,j-1}$ depending on the state of $bt[i][j]$.

If at any time, we move in some direction that contradicts the movement that the ball would have had if it landed on $b_{i,j}$ in the original input, we add $b_{i,j}$ to the list of flipped boards, which we output after we are done backtracking.