Let $P$ be a point in $\mathbb{P}^n$. Then we have a regular map $\mathbb{P}^n\setminus P\rightarrow \mathbb{P}^{n-1}$, which is not defined only at $P$. Then We know that the graph closure of this map is the Blow Up of $\mathbb{P}^n$ at $P$. $Blow_P \mathbb{P}^n\hookrightarrow \mathbb{P}^n\times \mathbb{P}^{n-1}$. Now Let $p_2$ be the projection to the second factor. $p_2:Blow_P \mathbb{P}^n\rightarrow \mathbb{P}^{n-1}$. My question is what is $p_2^*\mathcal{O}(1)$? Is it $\mathcal{O}(-E)$ or $p_1^*\mathcal{O}(1)\otimes \mathcal{O}(-E)$?

Let $P$ be a point in $\mathbb{P}^n$. Then we have a regular map $\mathbb{P}^n\setminus P\rightarrow \mathbb{P}^{n-1}$, which is not defined only at $P$. Then We know that the graph closure of this map is the Blow Up of $\mathbb{P}^n$ at $P$. $Blow_P \mathbb{P}^n\hookrightarrow \mathbb{P}^n\times \mathbb{P}^{n-1}$. Now Let $p_2$ be the projection to the second factor. $p_2:Blow_P \mathbb{P}^n\rightarrow \mathbb{P}^{n-1}$. My question is what is $p_2^*\mathcal{O}(1)$? Is it $\mathcal{O}(-E)$ or $p_1^*\mathcal{O}(1)\otimes \mathcal{O}(-E)$?

Here $E$ is the exceptional divisor and $p_1$ is the projection onto the first factor.