An eigenvector of a linear tranformation $T\colon\mathbf{V}\to\mathbf{V}$ is a vector $\mathbf{v}$, $\mathbf{v}\neq\mathbf{0}$, such that there is a scalar $\lambda$ for which $T(\mathbf{v})=\lambda \mathbf{v}$. The scalar $\lambda$ is called an eigenvalue. $T$ is diagonalizable if and only if you can find a basis for $\mathbf{V}$ in which every vector in the basis is an eigenvector. This does not mean that any basis will consist of eigenvectors. It means that there is a way to pick a basis so that every element of the basis is an eigenvector.
So, here, $\mathbf{V}=\mathbf{P}_2$. Your function $T$ is defined by the given formula. For example, if $p(x) = x^2 - 3x + 1$, then $p(-x) = (-x)^2 -3(-x) + 1 = x^2 +3x + 1$, so $T(p) = x^2+3x+1$.
You are looking for eigenvalues and eigenvectors. If you already know about the characteristic polynomial, then you should figure out the characteristic polynomial of $T$ (say, using the standard basis to find a coordinate matrix for $T$) and proceed along those lines.
If you don't know about the characteristic polynomial, not all is lost. Suppose $p(x) = ax^2+bx+c$ is an eigenvector. That means that $a$, $b$, and $c$ are not all zero, and there is a scalar $\lambda$ such that $T(p)=\lambda p$. Since $T(p(x)) = ax^2 -bx +c$ and $\lambda p = \lambda ax^2 + \lambda bx + \lambda c$, the equation says that
$$ax^2 - bx + c = T(p) = \lambda p = \lambda ax^2 +\lambda bx + \lambda c.$$
For this to happen, you must have:
\begin{align*}
a &= \lambda a\\
-b &= \lambda b\\
c &=\lambda c
\end{align*}
and $a$, $b$, and $c$ not all zero. You want to find all values of $\lambda$ for which there are soluctions $a,b,c$ not all zero. This is not terribly hard to do. Try that, and we can take it from there; edit your question if you get stuck to account for how far you have managed to go then.
Let $v_1=(-2,1,0), v_2=(-2,1,1)$ and $v_3=(1,2,0)$ so $B=(v_1,v_2,v_3)$ is a basis of $\Bbb R^3$ and the two former vectors span the plane $x+2y=0$ and the last vector is orthogonal to it. The matrix of the reflection $T$ in the basis $B$ is
$$[T]_B=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$
and clearly that $1,1$ and $-1$ are the eigenvalues associated to the eigenvectors $v_1,v_2$ and $v_3$ respectively.
now let $P$ the change matrix from the standard basis $B_c$ to $B$ then
$$P=(v_1\;v_2\;v_3)$$
and then the matrix of $T$ in the standard basis is
$$[T]_{B_c}=P[T]_BP^{-1}=\frac15\begin{pmatrix}3&-4&0\\-4&-3&0\\0&0&5\end{pmatrix}$$
Best Answer
Finding the matrix representation is the surest way to find the eigenvalues and eigenvectors.
Hint: To start you need a basis for $P_1$. Any basis will do so I would choose the obvious one: $\{1, x\}$. Next to find the matrix $A$ that represents $T$ with respect to this basis you need to remember the following: The $i^\text{th}$ column of $A$ is the coordinate representation of $T(e_i)$ where $e_i$ is the $i^\text{th}$ element in your basis.
So the first and second columns of $A$ in your case are going to be the coordinate representations of $T(1)$ and $T(x)$ respectively.
Once you have written down $A$ you need to compute the characteristic polynomial and find it's roots to get the eigenvalues. Then for each eigenvalue $\lambda$ $$T(a + bx) = \lambda a + \lambda bx$$ gives you a system of two equations whose solution set is the eigenspace for $\lambda$.