There are $\binom{2n}{n+1}$ monotonic paths from $\langle 0,0\rangle$ to $\langle n-1,n+1\rangle$: such a path must contain exactly $(n-1)+(n+1)=2n$ steps, any $n+1$ of those steps can be vertical, and the path is completely determined once you know which $n+1$ of the $2n$ steps are vertical. Every monotonic path from $\langle 0,0\rangle$ to $\langle n-1,n+1$ necessarily rises above the diagonal, since it starts on the diagonal and finishes above it. At some point, therefore, it must go from a point $\langle m,m\rangle$ on the diagonal to the point $\langle m,m+1\rangle$ just above the diagonal. After the point $\langle m,m+1\rangle$ the path must still take $(n+1)-(m+1)=n-m$ vertical steps and $(n-1)-m=n-m-1$ horizontal steps.
If you flip that part of the path across the axis $y=x+1$, each vertical step turns into a horizontal step and vice versa, so you’re now taking $n-m$ horizontal and $n-m-1$ vertical steps. You’re starting at $\langle m,m+1\rangle$, so you end up at $\langle m+(n-m),(m+1)+(n-m-1)\rangle=\langle n,n\rangle$. Thus, each monotonic path from $\langle 0,0\rangle$ to $\langle n-1,n+1\rangle$ can be converted by this flipping procedure into a monotonic path from $\langle 0,0\rangle$ to $\langle n,n\rangle$ that has vertical step from some $\langle m,m\rangle$ on the diagonal to the point $\langle m,m+1\rangle$ immediately above it.
Conversely, every monotonic path from $\langle 0,0\rangle$ to $\langle n,n\rangle$ that rises above the diagonal must have such a vertical step in it, and reversing the flip produces a monotonic path from $\langle 0,0\rangle$ to $\langle n-1,n+1\rangle$. Thus, this flipping procedure yields a bijection between all monotonic paths from $\langle 0,0\rangle$ to $\langle n-1,n+1\rangle$, on the one hand, and all monotonic paths from $\langle 0,0\rangle$ to $\langle n,n\rangle$ that rise above the diagonal, on the other. As we saw in the first paragraph, there are $\binom{2n}{n+1}$ of the former, so there are also $\binom{2n}{n+1}$ of the latter. The total number of monotonic paths from $\langle 0,0\rangle$ to $\langle n,n\rangle$, on the other hand, is $\binom{2n}n$: each path has $2n$ steps, any $n$ of them can be vertical, and the path is completely determined once we know which $n$ are vertical.
The difference $\binom{2n}n-\binom{2n}{n+1}$ is therefore simply the total number of monotonic paths from $\langle 0,0\rangle$ to $\langle n,n\rangle$ minus the number that rise above the diagonal, i.e., the number that do not rise above the diagonal — which is precisely what we wanted.
Best Answer
Sorry for pictures like that but...
Lets denote C as the point, where our good path touched diagonal the last time.
Lets put good paths to the sets according to their points $C(m,m)$. We have $C(0,0),\dots,C(n-1,n-1)$. How many paths belong to the set of point $C(m,m)$?
There are $P_m$ ways to go to the point $C(m,m)$. We know that for the rest part of the path we wont touch the diagonal anymore, so that means that our path lies under the line DE. Which means, that for the rest part of the path we have $P_{n-(m+1)}$ ways to do that. So the group of $C(m,m)$ consists of $P_m\cdot P_{n-(m+1)}$ good paths. We may take $m=0,\dots,n-1$ and then the sum of all possible paths is $\sum_{m=0}^{n-1}P_m\cdot P_{n-m-1}=P_n$