Answer
$12.8KN$
Work Step by Step
We know that there are four bolt used for the connection and therefore
$N=4T$
$N=4(4KN)=16KN$
As $F=\mu_s N$
$\implies F=(0.4)(16KN)=6.4KN$
We know that the sum of forces in the x-direction is zero
$\Sigma F_x=0$
$\implies P-2F=0$
$\implies P=2F$
We plug in the known values to obtain:
$P=2(6.4KN)=12.8KN$
