$A = \pi r^2 = \pi(50^2)$
$A = 2500\pi ~ \text{mm}^2$
$I = \dfrac{\pi r^4}{4} = \dfrac{\pi(50^4)}{4}$
$I = 1\,562\,500 \pi ~ \text{mm}^4$
$M = 250P ~ \text{N}\cdot\text{mm}$
Compression fibers are critical (fibers at B)
$\sigma_B = -\sigma_a - \sigma_f$
$\sigma_B = -\dfrac{P}{A} - \dfrac{Mr}{I}$
$-80 = -\dfrac{P}{2500\pi} - \dfrac{(250P)(50)}{1\,562\,500 \pi}$
$80 = \dfrac{P}{2500\pi} + \dfrac{P}{125\pi}$
$80\pi = \dfrac{P}{2500} + \dfrac{P}{125}$
$80\pi = \dfrac{21}{2500}P$
$P = 29,919.9 ~ \text{N}$
$P = 29.92 ~ \text{kN}$
Check for fibers at A
$\sigma_A = -\sigma_a + \sigma_f$
$\sigma_A = -\dfrac{P}{A} + \dfrac{Mr}{I}$
$\sigma_A = -\dfrac{29\,919.9}{2500\pi} + \dfrac{(250 \times 29\,919.9)(50)}{1\,562\,500 \pi}$
$\sigma_A = 72.38 ~ \text{MPa} \lt 80 ~ \text{MPa}$ (okay)
Hence,
$P = 29.92 ~ \text{kN}$ answer
