Number of parenthesizations and Catalan numbers
Define function $Q:\mathbb N\to\mathbb N$  such that $Q(n)=P(n+1)$ for all $n\ge 0$. Or, what is equivalent, $P(n)=Q(n-1)$ for all $n\ge 1$.

We have $$Q(0)= P(0+1)=P(1)=1.\quad\quad\quad\quad\quad\quad\quad\quad$$

Also, for $n\ge0$,
$$\begin{aligned}
  Q(n+1)&amp;=P(n+2)\\
   &amp;=\sum^{(n+2)-1}_{k=1} P(k)P((n+2)-k)\\
   &amp;=\sum^{n+1}_{k=1} Q(k-1)Q(n+1-k)\\
   &amp;\stackrel{k=i+1}{=\!=\!=\!=}\sum^{n}_{i=0} Q((i+1)-1)Q(n+1-(i+1))\\
   &amp;=\sum^{n}_{i=0} Q(i)Q(n-i).\\
\end{aligned}$$

We see that $Q(\cdot)$ and the Catalan numbers $C(\cdot)$ share the same initial condition and the same recurrence relation. So, we have, for all $n\ge0$,
$$ Q(n) = C(n).$$
So, for all $n\ge 1,$
$$P(n)=Q(n-1)=C(n-1)={\tfrac {1}{n}}{\tbinom {2(n-1)}{n-1}}$$

Exercise. Define function $D$ such that $D(n)=C(n-1)$ for $n\ge1$. Show that $D(n)=P(n)$ for $n\ge1$, assuming that you have not seen the deduction above.