There are two primary viewpoints for the BJT: the DC operating point and the AC behavior around that DC operating point. You must keep these ideas totally separate in your mind. The connection between them is simply that the DC operating point sets the value of the tangent line around which the BJT operates when a tiny AC signal is applied. But the AC behavior has nothing to do with the DC operating point. Nothing! So get that in mind.
Let's look at your schematic this way:
The first thing I can say about this is that the BJT cannot saturate. It will always be in active mode. The reason is that there is no way for the BC-junction to become forward-biased in this circuit. The base voltage cannot be higher than \$9\:\text{V}\$. Just not possible here. So the BJT is in active mode. And this means that the concept of \$\beta\$ applies well.
I don't even have to ask the question. The BJT is active, not saturated. Said and done.
This makes it very easy to work out, using KVL, all the details for the DC operating point of the circuit -- in theory. Of course, any specific physical BJT used in a real circuit will have varying details -- the \$\beta\$ can vary fairly widely, for example. (So can the saturation current.) But at least, in theory, we can write the following:
$$V_{_\text{TH}} -I_{_\text{B}}\cdot R_{_\text{TH}} -V_{_\text{BE}}-I_{_\text{E}}\cdot R_1=0\:\text{V}$$
where \$V_{_\text{TH}}=9\:\text{V}\cdot\frac{R_3}{R_2+R_3}\$ and \$R_{_\text{TH}}=\frac{R_2\,\cdot\,R_3}{R_2+R_3}\$.
Since you know the BJT is in active mode -- and I cannot emphasize it enough that you should be able to strongly agree with me on this point -- then as I said \$\beta\$ applies and we can say that \$I_{_\text{E}}=I_{_\text{B}}\cdot \left(\beta+1\right)\$. That allows us to substitute into the above equation, re-arrange, and find:
$$I_{_\text{B}}=\frac{V_{_\text{TH}}-V_{_\text{BE}}}{R_{_\text{TH}}+\left(\beta+1\right)\cdot\,R_1}$$
You will find that Spice programs will reach a value close to that, assuming you look at the BJT model details being applied by the program.
From the above, you can find all of the node voltages for the DC operating point.
Now, let's get to your updated portion of your question, given at the bottom.
None of that disagrees with the KVL I applied earlier. So it is consistent.