= Solution
Use the <characteristic polynomial> convention $p(\lambda)=\det(\lambda I-P)$, with $I$ the <identity matrix>. Put $u=\lambda-1/2$. Then
$$
\begin{aligned}
p(\lambda)&=\det\begin{pmatrix}u&-1/2&-1/2\\-1/4&u&0\\-1/4&0&u\end{pmatrix}\\
&=u^3-\frac18u-\frac18u=u\left(u^2-\frac14\right)\\
&=\lambda(\lambda-1/2)(\lambda-1).
\end{aligned}
$$
The required <characteristic polynomial> equation for this <matrix> is therefore
$$
\boxed{p(\lambda)=\lambda^3-\frac32\lambda^2+\frac12\lambda=0.}
$$
Using $\det(P-\lambda I)$ instead negates the cubic <polynomial> and leaves the <eigenvalues> unchanged.
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