I'm wondering if we may count multiple roots? For example if $f(x)-a=(x-x_0)^2g(x)$ , should $x_0$ be counted twice ?

The problem just says "real roots", so I think that's possible to count multiple roots.

I have a rather "weird" proof: Let such polynomial exists. Draw the graph of the polynomial; let $A_1, A_2, \cdots, A_n$ be the turning points of the polynomial (those $x$ where $f'(x) = 0$. If the polynomial has an odd degree, for a sufficiently large $a$, $f(x)=a$ only has one solution. So the polynomial must have an even degree, thus $n$ is odd. (Count multiple turning points as multiple occurences.) WLOG assume the polynomial has a positive leading coefficient; that is, the polynomial opens upwards. The trick is to note that however curvy the curve between two points is, as long as there is no turning point between the two points, the curve can be replaced by a straight line, and it will not be horizontal (otherwise there is a turning point between them). Moreover, we can assume that for a sufficiently large $a$, no part of the curve between $A_1$ and $A_n$ goes above the line $y=a$, so there are only two solutions for this $a$. Cut the portion above the curve (since it also has two solutions for each $a$), and denote the intersections between the curve and the line $y=a$ be $A_0$ and $A_{n+1}$. After this, replace all curves with lines adjoining two points $A_iA_j$ where $j-i=1$. Let $A_p$ be one of the points that has the minimum $y$-value. If no other $A_i$ has the same minimum value, then if $P$ is the $y$-value of $A_p$, $f(x) = P$ only has one solution. So there is at least another $A_i$ with the same minimum. Now, note that if we remove $A_p$ and join $A_{p-1}$ and $A_{p+1}$, the number of solutions of $f(x)=a$ for each $a$ stays the same parity except at $P$, where it decreases by 1 and thus changing its parity. So if there are two points with the same $y$-value, removing both of them causes the number of solutions of $f(x)=a$ for each $a$ to stay with the same parity. However, there are only an odd number of points between $A_0$ and $A_{n+1}$, thus not all can be paired; the $y$-value of any point left out is the value of $a$ such that $f(x)=a$ has an odd number of solution. I know it's weird and hard to understand, but it's what I have managed to find. If I can rephrase it better to be more "mathematical", I'll post it.

what if $P(x)=x^{2}$ and we count mutiple roots when $a=0$ ?

From reading the problem (again), I think you don't count multiple roots as multiple (otherwise it's trivial as you mentioned).