Free body diagram:
For aluminum:
$\Sigma M_B = 0$
$6P_{al} = 2.5(50)$
$P_{al} = 20.83 \, \text{kN}$
$\delta = \dfrac{PL}{AE}$
$\delta_{al} = \dfrac{20.83(3)1000^2}{500(70\,000)}$
$\delta_{al} = 1.78 \, \text{mm}$
For steel:
$\Sigma M_A = 0$
$6P_{st} = 3.5(50)$
$P_{st} = 29.17 \, \text{kN}$
$\delta = \dfrac{PL}{AE}$
$\delta_{st} = \dfrac{29.17(4)1000^2}{300(200\,000)}$
$\delta_{st} = 1.94 \, \text{mm}$
Movement diagram:
$\dfrac{y}{3.5} = \dfrac{1.94 - 1.78}{6}$
$y = 0.09 \, \text{mm}$
$\delta_B = \, \text{vertical movement of } P$
$\delta_B = 1.78 + y = 1.78 + 0.09$
$\delta_B = 1.87 \, \text{mm}$ answer