The explanation for this assumption is the same for most assumptions: because it makes the problem easier. This equation is (generally) applied to a single streamline due to the assumptions that were made when the equation was derived. Many mistakenly ascribe Bernoulli's Law to the principle of conservation of energy, when in reality it is a direct consequence of Newton's linear momentum equation. From the relatively straightforward force analysis of a differential fluid mass, it can be shown that
$$-\frac{\partial p}{\partial s}=\rho a_s=\rho v\frac{\partial v}{\partial s},
\tag{i}$$
$$+\frac{\partial p}{\partial n}=\rho a_n=\rho\frac{v^2}{R}, \tag{ii}$$
where $p$ is the static pressure, $\rho$ is the fluid density, $a$ is the local acceleration, $v$ is velocity, $R$ is the local radius of curvature, and $s$ and $n$ are curvilinear coodinates along and normal to the streamline, respectively. A partial differential is used because the pressure and velocity (in general) change in both the $n$ and $s$ directions. Now, if we limit our analysis to changes only along the streamline, we can replace the original partial differentials in (1) with exact differentials. Rearranging, this gives us
$$\frac{dp}{ds}+\rho V\frac{dV}{ds}=0, \tag{iii}$$
which can be simplified further into the classic differential Bernoulli equation:
$$\frac{dp}{\rho}+VdV=0. \tag{iv}$$
It is this version of the equation (with it's inherent assumptions) that is then integrated to give the classic textbook version of Bernoulli's Eqn. mentioned earlier.
$$p+\frac{1}{2}\rho V^2=p_0$$
Why would we do this? Well, there are several flow situations in which it is approximately valid (e.g. irrotational flows), where the stagnation pressure is uniform everywhere and needs to be calculated only once. For viscous flows, the equation can still be used to determine the stagnation pressure at a given location in the flow, but there should be no expectation that the stagnation pressures will be equal between streamlines.
I wasn't able to figure out what you did, so here is my analysis, without the resistance. Let:
Q = Total volume flow rate
$Q_a$ = Volume flow rate into converging pipe
$Q_b$ = Volume flow rate into diverging pipe
$p_1a$ = static pressure just after entrance to a
$p_2a$ = static pressure just before exit from a
$p_1b$ = static pressure just after entrance to a
$p_2b$ = static pressure just before exit from a
$T_1$ = "total pressure" in channel leading up to diffluence
$T_2$ = "total pressure" in channel after diffluence
$A_{a1}$ = cross sectional area of converging pipe at inlet
$A_{a2}$ = cross sectional area of converging pipe at outlet
$A_{b1}$ = cross sectional area of diverging pipe at inlet
$A_{b2}$ = cross sectional area of diverging pipe at outlet
CASE OF NO FRICTIONAL LOSS
Bernoulli equations relevant to pipe a:
$$T_1=p_1+\rho \frac{(Q_a/A_{a1})^2}{2}$$
$$p_1+\rho \frac{(Q_a/A_{a1})^2}{2}=p_2+\rho \frac{(Q_a/A_{a2})^2}{2}$$
$$p_2+\rho \frac{(Q_a/A_{a2})^2}{2}=T_2$$
Adding these three equations together gives $$T_1=T_2$$
Thus, for the case without friction, energy is conserved and the "total pressure" after the split section is equal to the "total pressure" before the split section. This is irrespective of how the flow splits between the two sections. The Bernoulli equations for pipe b will give the same result. Also, the convergence and divergence in the channels doesn't matter, as long as the final outlet pipe has the same cross sectional area as the initial inlet pipe.
CASE WITH FRICTIONAL EFFECTS INCLUDED
Bernoulli equations relevant to pipe a:
$$T_1=p_1+\rho \frac{(Q_a/A_{a1})^2}{2}$$
$$p_1+\rho \frac{(Q_a/A_{a1})^2}{2}=p_2+\rho \frac{(Q_a/A_{a2})^2}{2}+k_a\rho \frac{(Q_a/A_{a1})^2}{2}$$
$$p_2+\rho \frac{(Q_a/A_{a2})^2}{2}=T_2$$
Adding these three equations together gives $$T_1=T_2+k_a\rho \frac{(Q_a/A_{a1})^2}{2}\tag{1}$$
Similarly, for channel b:$$T_1=T_2+k_b\rho \frac{(Q_b/A_{b1})^2}{2}\tag{2}$$
Thus, for the case with friction, mechanical energy is not conserved and the "total pressure" after the split section is not equal to the "total pressure" before the split section. Moreover, the split between the two channels is relevant.
Mass balance equation:
$$Q_a+Q_b=Q\tag{3}$$
Eqns. 1-3 provide three algebraic equations in the three unknowns $(T_1-T_2)$, $Q_a$, and $Q_b$.
Best Answer
I'm assuming incompressible fluid here and rigid pipes (no compliance).
Before the actual physical split is there any appreciable resistance compared to RB and RA? If not then PA = PB = 19" wg.
If there is a significant resistance you need to include it in the model as Rin, and then
PA = PB = 19" *(RA || RB)/(Rin + RA || RB)
In any event PA = PB