= Solution
For $\mathbf1=(1,1,1)^T$, the entries of $P^T\mathbf1$ are the column sums of $P$. Hence
$$
P^T\mathbf1=\begin{pmatrix}1\\1\\1\end{pmatrix}=\mathbf1,
$$
so $1$ is an <eigenvalue> of $P^T$, with <eigenvector> $\mathbf1$.
The <determinant> is unchanged by <matrix transposition>, so
$$
\det(P-I)=\det((P-I)^T)=\det(P^T-I)=0.
$$
Thus $P-I$ has a nontrivial <null space>, giving
$$
\boxed{1\text{ is also an eigenvalue of }P.}
$$
The same argument with $\lambda I$ shows that a <matrix> and its transpose have the same <characteristic polynomial>. It does not say that they have the same <eigenvectors>: here $P\mathbf1=(0.9,1.2,0.9)^T\ne\mathbf1$.
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