Given matrix P = [[2, 1], [-3, 0]], calculate P2 - 2P + 4I.

The matrix $P$ is given as:

$$P = \begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix}$$

The $2 \times 2$ unit matrix $I$ is:

$$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

$$P^2 = P \times P = \begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix}$$

Perform matrix multiplication:

$$P^2 = \begin{pmatrix} (2)(2) + (1)(-3) & (2)(1) + (1)(0) \\ (-3)(2) + (0)(-3) & (-3)(1) + (0)(0) \end{pmatrix}$$ $$P^2 = \begin{pmatrix} 4 - 3 & 2 + 0 \\ -6 + 0 & -3 + 0 \end{pmatrix}$$ $$P^2 = \begin{pmatrix} 1 & 2 \\ -6 & -3 \end{pmatrix}$$

Multiply each element of matrix $P$ by 2:

$$2P = 2 \begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix} = \begin{pmatrix} 2 \times 2 & 2 \times 1 \\ 2 \times (-3) & 2 \times 0 \end{pmatrix}$$ $$2P = \begin{pmatrix} 4 & 2 \\ -6 & 0 \end{pmatrix}$$

Multiply each element of matrix $I$ by 4:

$$4I = 4 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 \times 1 & 4 \times 0 \\ 4 \times 0 & 4 \times 1 \end{pmatrix}$$ $$4I = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}$$

Substitute the calculated matrices into the expression:

$$P^2 - 2P + 4I = \begin{pmatrix} 1 & 2 \\ -6 & -3 \end{pmatrix} - \begin{pmatrix} 4 & 2 \\ -6 & 0 \end{pmatrix} + \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}$$

First, perform the subtraction:

$$\begin{pmatrix} 1 & 2 \\ -6 & -3 \end{pmatrix} - \begin{pmatrix} 4 & 2 \\ -6 & 0 \end{pmatrix} = \begin{pmatrix} 1 - 4 & 2 - 2 \\ -6 - (-6) & -3 - 0 \end{pmatrix}$$ $$= \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}$$

Now, perform the addition:

$$\begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} + \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} -3 + 4 & 0 + 0 \\ 0 + 0 & -3 + 4 \end{pmatrix}$$ $$= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

The result is the $2 \times 2$ unit matrix. Comparing this to the given options, it matches option C.

The final answer is $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.