Without the counter diagonals (in reference to the figure below)
$\Sigma F_v = \frac{2}{3}P - P = -\frac{1}{3}P$
$\Sigma F_v = \frac{1}{3}P ~ \text{downward}$
Upward force is needed to put the downward P/3 into equilibrium. The vertical component of a diagonal will provide the equilibrant force. Since the needed equilibrant is upward, BE will be in compression, thus, redundant. The acting member (diagonal in tension) therefore is CD.
$\Sigma M_C = 0$
$\frac{2}{3}P(L) = 20L$
$\frac{2}{3}P = 20$
$P = 30 ~ \text{kN}$ answer
$\Sigma F_V = 0$
$F_{CD} \sin 45^\circ + \frac{2}{3}P = P$
$F_{CD} \sin 45^\circ = \frac{1}{3}P$
$F_{CD} \sin 45^\circ = \frac{1}{3}(30)$
$F_{CD} = 10\sqrt{2} ~ \text{kN} = 14.14 ~ \text{kN}$ answer