Without the loads P:
$\tau = \dfrac{T\rho}{J}$

Where:

$T = 14(10) = 140 \, \text{kip}\cdot\text{in}$
$\rho = \sqrt{13} \, \text{in}$

$J = \Sigma A\rho^2 = \frac{1}{4}\pi (\frac{7}{8})^2 \, [ \, 4(\sqrt{13})^2 + 2(2)^2 \, ] = 36.08 \, \text{in}^4$

$\tau_{maximum} = \dfrac{140\sqrt{13}}{36.08} = 14.0 \, \text{ ksi}$ *answer*

$\tau_{minimum} = \dfrac{140(2)}{36.08} = 7.76 \, \text{ ksi}$ *answer*

Note:

The τ_{maximum} is carried by the corner rivets (4 rivets in all) while the τ_{minimum} is carried by the middle two rivets.

With the loads P, two cases will arise:

1st case (*P 14 kips*)

$T = 10(14) - 6P = (140 - 6P) \, \text{kip}\cdot\text{in}$
$\tau = \dfrac{T \rho}{J}$

$8000 = \dfrac{(140 - 6P)(1000)(\sqrt{13})}{36.08}$

$80.05 = 140 - 6P$

$P = 10.0 \, \text{ kips}$ *answer*

Note:

Without the load P = 10 kips, the shear stress at corner rivets is 14 ksi (see τ_{maximum}), which is way above 8 ksi. Thus, this minimum value of P is necessary to prevent stressing the corner rivets beyond 8 ksi.

2nd case (*P > 14 kips*)

$T = 6P - 10(14) = (6P - 140) \, \text{kip}\cdot\text{in}$

$\tau = \dfrac{T \rho}{J}$

$8000 = \dfrac{(6P - 140)(1000)(\sqrt{13})}{36.08}$

$80.05 = 6P - 140$

$P = 36.68 \, \text{kips}$ *answer*