= Solution
First let $B_{ab}=\nabla_bU_a$ without projection. Commute derivatives on this <covector> and use affine <geodesic> motion:
$$
U^c\nabla_cB_{ab}=\nabla_b(U^c\nabla_cU_a)-(\nabla_bU^c)(\nabla_cU_a)-U^cU^dR_{dacb}=-B_{ac}B^c{}_b-R_{cadb}U^cU^d.
$$
The last equality relabels the two contracted null-vector indices. Nullness and affine parametrization give $U^aB_{ab}=0$ and $B_{ab}U^b=0$. Therefore inserting $P$ into the contracted middle index changes no screen-projected quadratic term. Since $\nabla_UP=0$, screen projection proves the <Sachs optical equations>
$$
\boxed{U\cdot\nabla\widehat B_{ab}=-\widehat B_{ac}\widehat B^c{}_b-\mathcal R_{ab},\qquad\mathcal R_{ab}=P_a{}^eP_b{}^fR_{cedf}U^cU^d.}
$$
The optical tidal <matrix> $\mathcal R_{ab}$ is symmetric by the <Riemann curvature tensor>'s pair symmetries. Raising one screen index gives the corresponding equation for $\widehat B^a{}_b$.
In terms of the defined optical scalars, the equation is explicitly
$$
\begin{aligned}
U\cdot\nabla\widehat B_{ab}={}&-\frac{\theta^2}{4}P_{ab}-\theta(\widehat\sigma_{ab}+\widehat\omega_{ab})\\
&-\widehat\sigma_{ac}\widehat\sigma^c{}_b-\widehat\omega_{ac}\widehat\omega^c{}_b\\
&-\widehat\sigma_{ac}\widehat\omega^c{}_b-\widehat\omega_{ac}\widehat\sigma^c{}_b-\mathcal R_{ab}.
\end{aligned}
$$
This <matrix> identity contains the <null expansion>, <null shear> and <null twist> evolution equations as its trace, symmetric trace-free part and antisymmetric part.
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