Answer
see the details below.
Work Step by Step
Consider the set of vectors $\{1,x,x^2,x^3\}$ in $P_3$ with the inner product
$$\langle p,q\rangle=a_0b_0+a_1b_1+a_2b_2+a_3b_3.$$
Now, we have
$$\langle 1,x\rangle=0, \quad \langle 1,x^2\rangle=0, \langle 1,x^3\rangle=0,$$
$$\langle x, x^2\rangle=0, \quad \langle x,x^3\rangle=0, \langle x^2,x^3\rangle=0,$$
hence the set is orthogonal. Also, we have
$$\| 1\|=1, \quad \| x\|=1 , \quad \| x^2\|=1, \quad\| x^3\|=1,$$
then the set is orthonormal.
Since the set is orthogonal and has 4 vectors and $P_3$ has the dimension $4$, then the set is a basis for $P_3$.
