$A = \pi r^2 = \pi(50^2)$

$A = 2500\pi ~ \text{mm}^2$

$I = \dfrac{\pi r^4}{4} = \dfrac{\pi(50^4)}{4}$

$I = 1\,562\,500 \pi ~ \text{mm}^4$

$M = 250P ~ \text{N}\cdot\text{mm}$

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*