Answer
$S$ is a basis for $P_3$.
Work Step by Step
The set $S=\left\{4 t-t^{2}, 5+t^{3}, 3 t+5,2 t^{3}-3 t^{2}\right\}$ is a basis for $P_3$. Indeed, consider the following linear combination
$$a(4 t-t^{2})+b( 5+t^{3})+c(3 t+5)+d ( 2 t^{3}-3 t^{2})=0.$$
Comparing the coefficients, we have
\begin{align*}
5b+5c&=0\\
4a+3c&=0\\
-a-3d&=0\\
b+2d&=0.
\end{align*}
The augmented matrix of the above system is given by
$$\left[\begin{array}{cccc}
{0}&{5}&{5}&{0}\\
{4}&{0}&{3}&{0}\\
{-1}&{0}&{0}&{-3}\\
{0}&{1}&{0}&{2}\\
\end{array}\right]$$
and its determinant is $30$. Therefore, the above system has unique solution, that is, the trivial solution. Then $S$ is linearly independent set of vectors and since $P_3$ has dimension $4$, then by Theorem 4.12, $S$ is a basis for $P_3$.
