= Solution
Taking the <matrix trace> of those matrices gives
$$
\begin{array}{c|rrrrr}
\text{class}&E&2S_4&C_2&2C_2^\prime&2\sigma_d\\\hline
\chi_p&3&-1&-1&-1&1
\end{array}
$$
The one-dimensional $p_z$ subspace has characters $(1,-1,1,-1,1)$, identifying $B_2$. The pair $(p_x,p_y)$ has characters $(2,0,-2,0,0)$, identifying $E$. Therefore
$$
\boxed{\Gamma_{p_x,p_y,p_z}=B_2\oplus E.}
$$
The dimensions add to three, as required. The $p_x,p_y$ pair must be retained together: a ninety-degree improper rotation mixes them, so either one alone is not a complete symmetry basis.<Tables for Group Theory by P. W. Atkins and others>[Printed page 12, $D_{2d}$ vector functions]{ref}
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