$P(1)$ is not simple:

To see why, consider the strange, type I, classical, simple, complex, LS $P(n)$, $n\geq 2$ realized as the set of complex, $(2n+2)\times(2n+2)$ matrices $\mathbf{M}$, with grading partitioning:
$$
\mathbf{M}=\begin{bmatrix}
\mathbf{A} & \mathbf{B} \\
\mathbf{C} & -\mathbf{A}
\end{bmatrix}
$$
where $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ are complex $(n+1)\times(n+1)$ matrices, with $tr\mathbf{A}=0$, $\mathbf{B}$ symmetric and $\mathbf{C}$ antisymmetric. The dimensions of the odd and the even part are $n(n+2)$ and $(n+1)^2$ respectively.

Thus, the even subalgebra is $A_n=sl_{n+1}$ i.e. the set of complex, traceless matrices $\mathbf{A}$.

Now for $n=1$, the above definition makes sense however, the resulting LS $P(1)$ is not simple, because the basis elements of the even part together with $\mathbf{e}_{13}$ (that is the matrix of suitable dimension with a sole non-zero entry: $\big(\mathbf{e}_{13}\big)_{ij}=\delta_{1i}\delta_{3j}$), form the basis of a proper, graded ideal.