$\Sigma M_B = 0$

$6P = 2T \sin 10^\circ$

$3P = T \sin 10^\circ$ → Equation (1)

$\Sigma F_H = 0$

$B_H = T \cos 10^\circ$

From Equation (1),

$T = \dfrac{3P}{\sin 10^\circ}$

Thus,

$B_H = \left( \dfrac{3P}{\sin 10^\circ} \right) \cos 10^\circ$

$B_H = 3 \cot 10^\circ \, P$

$\Sigma F_V = 0$

$B_V = T \sin 10^\circ + P$

Again from Equation (1),

$T \sin 10^\circ = 3P$

Thus,

$B_V = 3P + P$

$B_V = 4P$

${R_B}^2 = {B_H}^2 + {B_V}^2$

${R_B}^2 = (3 \cot 10^\circ \, P)^2 + (4P)^2$

$R_B^2 = 305.47P^2$

$R_B = 17.48P$

$P = \dfrac{R_B}{17.48}$ → Equation (2)

Based on tension of rod (equation 1):

$P = \frac{1}{3} T \sin 10^\circ$

$P = \frac{1}{3} [ \, 5000 \times \frac{1}{4} \pi (0.5)^2 \, ] \sin 10^\circ$

$P = 56.83 \, \text{lb}$

Based on shear of rivet (equation 2):

$P = \dfrac{4000 [ \, \frac{1}{4} \pi (0.25)^2 \, ]}{17.48}$

$P = 11.23 \, \text{lb}$

Safe load P,

$P = 11.23 \, \text{ lb}$ *answer*