$M_1 = 0$
$M_2 = -Pb$
$M_3 = 0$
$h_1 = 0$
$| \, h_3 \, | = \delta_3$
$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right)$
$0 + 2(-Pb)(a + b) + 0 + 0 + 0 = 6EI\left( 0 + \dfrac{h_3}{b} \right)$
$-2PbL = \dfrac{6EI \, h_3}{b}$
$EI \, h_3 = -2PLb^2 / 3$ ← the negative sign indicates that point 3 is below point 2
Thus,
$EI \, \delta_3 = PLb^2 / 3$ answer
If P is replaced by a clockwise couple M
$M_1 = 0$
$M_2 = -M$
$M_3 = -M$
$h_1 = 0$
$| \, h_3 \, | = \delta$
$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right)$
$0 + 2(-M)(a + b) + (-M)(b) + 0 + 0 = 6EI\left( 0 + \dfrac{h_3}{b} \right)$
$-2ML - Mb = \dfrac{6EI \, h_3}{b}$
$-M(b + 2L) = \dfrac{6EI \, h_3}{b}$
$-Mb(b + 2L) = 6EI \, h_3$
$EI \, h_3 = -Mb(b + 2L) / 6$ ← point 3 is below point 2
Thus,
$EI \, \delta_3 = Mb(b + 2L) / 6$ answer
