Let $X\subseteq\mathbb{P}^n\times\mathbb{P}^m\subseteq \mathbb P^{(n+1)(m+1)-1}$ be a projective variety of dimension $p$ and degree $d$ defined over an algebraically closed field $k$, where $\mathbb{P}^n\times\mathbb{P}^m\subseteq \mathbb P^{(n+1)(m+1)-1}$ via the Segre embedding. Let $X'\subseteq \mathbb{P}^n$ be the projection of $X$ on $\mathbb{P}^n$, which we will assume having dimension equal to that of $X$. What can we say about the degree $d'$ of $X'$? Is there some formula linking the degrees $d$ and $d'$. Since an analytical charcaterization is also good for my purposes, feel free to assume $k=\mathbb C$.

As abx says in his comment, all you can say is $d' \le d$. Here are some explicit examples with $m = 2$ and $n = 1$.

(1) To see that $d'$ can be arbitrarily smaller than $d$, fix a line $L \subseteq \mathbb{P}^2$ and set $X = L \times S$, where $S \subseteq \mathbb{P}^1$ is a finite set of points. Then $X$ has degree $\lvert S\rvert$ while has $X' = L$ has degree 1.

(2) To see that we can also have $d' = d$, take a degree $d$ curve $C \subseteq \mathbb{P}^2$ and set $X = C \times \{*\}$ for some point $* \in \mathbb{P}^1$. Then $X$ has degree $d$, and $X' = C$ also has degree $d$.

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