After $\mathrm{P}$ has been displaced $1 \mathrm{mm}$ downward, the angle at $\mathrm{P}$ is $\mathrm{\alpha }+\mathrm{\beta }=1.046365 \mathrm{rad}+0.522767 \mathrm{rad}=$ $1.569132 \mathrm{rad.}$
Set this angle equal to $\mathrm{\pi }/2-\mathrm{\gamma }$ and compute the shear strain $\mathrm{\gamma }$:

$$\begin{gathered} \mathrm{\pi }/2-\mathrm{\gamma }=1.569132 \mathrm{rad} \\ \therefore \mathrm{\gamma }=\mathrm{\pi }/2-1.569132 \mathrm{rad}=\mathbf{0}\mathbf{.}\mathbf{001664}\mathbf{ }\mathbf{rad} \end{gathered}$$