Classic Solution

Normal forces on characteristic intervals \begin{aligned} &N_{AB}=F\\ &N_{BC}=F-P\\ &N_{CD}=F-P+2P=F+P\\ \end{aligned} I express one cross-sectional area in terms of another to simplify further calculations \begin{aligned} &\frac{A_{AB}}{A_{BC}}=\frac{\frac{\pi\cdot d_2^2}{4}}{\frac{\pi\cdot d_1^2}{4}}=\frac{d_2^2}{d_1^2}=\frac{9}{16}=0.5625\\ &A_{AB}=0.5625\cdot A_{BC}\\ \end{aligned} I solve the condition from the task - the total elongation has to be equal to zero \begin{aligned} &\Delta_{l_C}=\Delta_{l_{AB}}+\Delta_{l_{BC}}+\Delta_{l_{CD}}\\ &\Delta_l=\frac{N\cdot l}{E\cdot A}\\ \end{aligned} \begin{aligned}\\ &\frac{F\cdot 4}{E\cdot 0.5625\cdot A_{BC}} + \frac{(F-P)\cdot 3}{E\cdot A_{BC}} + \frac{(F+P)\cdot 1}{E\cdot A_{BC}}=0 \ \ \ |\cdot EA_{BC}\\ &\frac{4}{0.5625}F + 3F - 3P + F + P=0\\ &11.11F=2P\\ &F=0.18\cdot 15\cdot 10^{3}\\ &F=2.7 \ kN\\ \end{aligned} Verification \begin{aligned} &\frac{2.6\cdot 10^{3}\cdot 4}{2.1\cdot 10^{11}\cdot\frac{\pi\cdot 0.15^{2}}{4}} + \frac{-12.3\cdot 10^{3}\cdot 3}{2.1\cdot 10^{11}\cdot\frac{\pi\cdot 0.2^{2}}{4}}+\frac{17.7\cdot 10^{3}\cdot 1}{2.1\cdot 10^{11}\cdot\frac{\pi\cdot 0.2^{2}}{4}}=6.17\cdot 10^{-9} \approx 0\\ \end{aligned}