Degree of $P$ as a smooth function equals its polynomial degree $d$. Let $M,N$ be connected oriented manifolds such that $\partial M=\partial N=\emptyset$.
Let $F:M\to N $ be a smooth proper map (i.e. for every $K\subset N$ compact, $F^{-1}(K)$ is compact). We define the degree of $F$ as
$$
deg(F):=\sum_{p\in F^{-1}(q)}\varepsilon(p)
$$
where $q$ is a regular value of $F$ and 
$$
\varepsilon(p)=\begin{cases}
   +1       & \quad \text{if F preserves the orientation near $p$ }\\
    -1  & \quad \text{if F reverves the orientation near $p$}\\
  \end{cases}
$$
Let $P$ be a complex polynomial of degree $d$. Considering $P$ as a map from $\mathbb{R}^2$ to $\mathbb{R}^2$, I proved that $P$ is smooth, proper (if and only if is polynomial degree $d$ is positive). 
I want to show that the degree of $P$ as a smooth function equals its degree as a polynomial.
(Somehow I think the fundamental theorem of algebra should be involved)
Any hints?
 A: I suppose you have shown that the degree does not depends on the regular value $q$ chosen (or the degree is not yet defined). 
First we show that $P$ is always orientation-preserving. Indeed, this is true for any holomorphic function: using the Cauchy Riemann equation, 
$$\begin{cases}
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \\
\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}.
\end{cases}$$
We have 
$$\det DP = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x} =\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 \ge 0$$
Thus $P$ is orientation-preserving (One can indeed show that the map is conformal).
Now pick $q\in \mathbb C$, then by the fundamental theorem of algebra, 
$$P(z) - q = C (z- x_1)^{n_1} \cdots (z-x_k)^{n_k}, $$
where $C\neq 0$, $x_i$'s are distinct and $n_1 + \cdots n_k = d$. Then 
$$P^{-1}(q) = \{x_1, x_2, \cdots, x_k\}.$$
Now if $q$ is a regular value of $P$, $DP(x_i)\neq 0$ for $i=1, \cdots k$. As $DP = D(P-q)$, this is the same as saying that $n_1 = n_2 = \cdots = n_k =1$. This implies that 
$k =d$, and $P^{-1} (q)$ has exactly $d$ point. As $P$ is orientation-preserving, we have $\deg (P) = d$. 
