Since the point D has no vertical movement, the net deformation due to forces acting on the bar must be zero.
Deformation of DC [Figure 2.33(b)]

d l_{ CD }=\frac{9 \times 1000 \times 300}{250 \times 200 \times 10^3}=0.054  mm

Deformation of CB (Figure 2.33(c))

d l_{ CB }=\frac{24 \times 1000 \times 200}{250 \times 200 \times 10^3}=0.096  mm

Deformation of BA (Figure 2.33(d))
In order to nullify the elongations of the portions DC and CB, the portion BA should be in compression, therefore, it is subjected to compressive force given by (24 – P) kN.

Therefore,  d l_{ BA }=\frac{(24-P) 10^3 \times 500}{250 \times 200 \times 10^3}=0.01(24-P)  mm

In order that the point D should not move

0.054 + 0.096 + 0.01(24 – P) = 0

P = 39 kN