When does a morphism preserve the degree of curves? Suppose $X \subset \mathbb{P}_k^n$ is a smooth, projective curve over an algebraically closed field $k$ of degree $d$ . In this case, degree of $X$ is defined as the leading coefficient of $P_X$, where $P_X$ is the Hilbert polynomial of $X$. 
I guess Hartshorne using following fact without proof: 
Under certain condition, the projection morphism $\phi$ from $O \notin X$ to some lower dimension space $\mathbb{P}_k^m$ gives a curve $\phi(X)$, and $deg X= deg\  \phi(X)$. I don't know which condition is needed to make this possible (the condition for $X \cong \phi(X)$ is given in the book), and more importantly, why they have the same degree.
One thing giving me trouble is the definition of degree by Hilbert polynomial. Is it possible to define it more geometrically?

Chris Dionne already explained the condition for $\phi$ to be an isomorphism from $X$ to $\phi(X)$. Suppose $\phi$ is a projection to $Q=\mathbb P^{n-1}$. Let $H'$ be a hyperplane in $Q$, then the schematic pre-image $\phi^{-1}Q=H\setminus \{ O\}$ where $H$ is a hyperplane of $P:=\mathbb P^n$ passing through $O$. Now 
$$O_P(H)|_X =(O_P(H)|_{P\setminus \{ O \}})|_X \simeq (\phi^*O_Q(H'))|_X \simeq (\phi|{}_X)^*(O_Q(H')|_{\phi(X)})$$
hence
$$ \deg X=\deg O_P(H)|_X= \deg O_Q(H')|_{\phi(X)}=\deg \phi(X).$$
