From the equation $ \begin{pmatrix} 4 & 3 \\ -1 & 0 \end{pmatrix} = \mathbf{P}^{-1} \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} \mathbf{P}, $ we can multiply both sides by $ \mathbf{P} $ on the left, to get \[\mathbf{P} \begin{pmatrix} 4 & 3 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} \mathbf{P}.\]Then \[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 4 & 3 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix},\]so \[\begin{pmatrix} 4a - b & 3a \\ 4c - d & 3c \end{pmatrix} = \begin{pmatrix} a & b \\ 3c & 3d \end{pmatrix}.\]Comparing entries, we get $ 4a - b = a, $ $ 3a = b, $ $ 4c - d = 3c, $ and $ 3c = 3d $. These reduce to $ b = 3a $ and $ c = d, $ so \[|a| + |b| + |c| + |d| = |a| + |3a| + |c| + |d| = 4|a| + 2|d|.\]Since $ a, $ $ b, $ $ c, $ and $ d $ are all nonzero integers, we minimize this by taking $ a = \pm 1 $ and $ d = \pm 1, $ in which case $ 4|a| + 2|d| = \boxed{6} $.