If $P\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$, then matrix P is equal to:

$P \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{2 \times 3} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}_{2 \times 3}$

Since the right-hand side is a $2\times3$ matrix and the matrix being multiplied by P is also $2\times3$, matrix P must be of order $2\times2$.

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Let:

$P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$

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From the first row, comparing columns on both sides:

$a + 4b = -7 \quad \cdots (i)$

$2a + 5b = -8 \quad \cdots (ii)$

$(i) \times 2: \quad 2a + 8b = -14 \quad \cdots (iii)$

$(iii) - (ii):$

$3b = -6$

$b = -2$

From $(i)$:

$a + 4(-2) = -7$

$a = 1$

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From the second row, comparing columns on both sides:

$c + 4d = 2 \quad \cdots (iv)$

$2c + 5d = 4 \quad \cdots (v)$

$(iv) \times 2: \quad 2c + 8d = 4 \quad \cdots (vi)$

$(vi) - (v):$

$3d = 0$

$d = 0$

From $(iv)$:

$c = 2$

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