confusion regarding bias voltage of a transistor
The circuit you have is a highly non-linear one, even if you assume that the AC voltage itself is exactly zero volts, peak to peak. Let's look at it without regard to the AC, for a moment. It'll be easier to show what I mean, then.



simulate this circuit &ndash; Schematic created using CircuitLab
I've removed the batteries, as they really just add &quot;noise&quot; to the diagram without helping at all in understanding it. It's enough to simply label the voltages, as they are implicitly referred to the ground reference.
Let's say that \$R_\text{C}=4.7\:\text{k}\Omega\$, \$V_\text{CC}=10\:\text{V}\$, \$V_\text{BB}=650\:\text{mV}\$ and that therefore \$I_\text{C}=1\:\text{mA}\$ (simplified Ebers-Moll model in active mode.) Then we'd expect to see a voltage drop of \$4.7\:\text{k}\Omega\cdot 1\:\text{mA}=4.7\:\text{V}\$ across \$R_\text{C}\$, meaning that the collector voltage sits at \$V_\text{C}=10\:\text{V}-4.7\:\text{V}=5.3\:\text{V}\$. All seems well, as the collector voltage is higher than the base voltage and so the BJT is really in active mode.
What happens if we change things so that \$V_\text{BB}=750\:\text{mV}\$? In this case, temporarily ignoring the collector resistor and assuming that the BJT stays in active mode, I'd expect the collector current to change by a factor of almost 50, to about \$I_\text{C}=50\:\text{mA}\$. (This is because there will be a factor of 10 change in the collector current for each and every increase of \$60\:\text{mV}\$ in the base-emitter voltage.) But now the voltage drop across \$R_\text{C}\$ should be nearly \$235\:\text{V}\$!! And that's just not possible. So already we know that even a very tiny adjustment of the base voltage leads to an impossible result -- if we assume the BJT is still in active mode.
Instead, what really happens is that the collector voltage will smash down close to the emitter voltage (about \$0\:\text{V}\$) and we will have a collector current nearing \$I_\text{C}=\frac{10\:\text{V}}{4.7\:\text{k}\Omega}\approx 2.13\:\text{mA}\$ and the BJT will be in saturation mode.
By now, I think you may be able to see why I hate that diagram so much. The BJT is a non-linear device. It responds wildly with respect to minor base-emitter voltage changes.
When your textbook says, &quot;This d.c. voltage is known as bias voltage and its magnitude is such that it always keeps the emitter-base junction forward biased regardless of the signal source...&quot;, it is just begging you to do exactly what you thought about doing -- adjust \$V_\text{BB}\$ so that no matter what the input signal AC peak magnitude might be the base-emitter junction remains in forward bias.
If you take their comment &quot;literally&quot; and apply it for any-sized AC input signal, then the entire idea becomes impossible! This is why you are asking a good question, right now.

When I was first trying to understand BJTs more than 50 years ago,
these same stupid examples and very similar writing about them caused
me similar confusion for a time. And I'm not happy to see that nothing
has changed in the education about BJTs in all that time.

Let's see why, using your own example of \$V_\text{AC}=5\:\text{V}_\text{PK}\$ for the AC source. You might then consider \$V_\text{BB}=6\:\text{V}\$, since that's large enough to keep the base-emitter junction forward-biased.
But that means that the base voltage is \$1\:\text{V} \le V_\text{B}\le 11\:\text{V}\$! But the collector current is a non-linear function of the base voltage (the emitter is grounded.) And I mean highly non-linear. For every \$60\:\text{mV}\$ change, the collector current is multiplied (or divided) by 10!! Look at that voltage swing, \$\Delta V_\text{B}=10\:\text{V}\$! How many orders of 10 in collector current does that imply??? About 170 orders of magnitude change!
Do you think there is any chance in this universe of that happening?
The impossible implications from taking their comment as being true under all circumstances you can imagine for the AC signal source means that you must let go of the idea that their model works for any imaginable AC signal magnitude. In short, you must instead imagine that it only applies for a very tiny AC signal magnitude. One that does NOT materially change the &quot;bias-point&quot; of the BJT. Or, put still another way, doesn't have a noticeable impact on the magnitude of \$V_\text{BB}\$.
In that case, the schematic they provide works.
But then you probably have another good question. Why did they bother to write, &quot;This d.c. voltage is known as bias voltage and its magnitude is such that it always keeps the emitter-base junction forward biased regardless of the signal source...&quot;? I mean, if the AC signal is assumed to be so tiny, what's the point of writing it that way?
Also, the way they write that almost makes you think that even if you keep the AC signal magnitude small, you might still be able to let the base-emitter junction swing widely if perhaps not as widely as you were imagining. For example, suppose \$V_\text{BB}=700\:\text{mV}\$. Then you might imagine that the peak AC voltage might be \$500\:\text{mV}\$. But that would NOT be valid for this circuit. Yet it meets with their statement, just fine, and you didn't even change \$V_\text{BB}\$!
So I think they chose poor writing.
Why not just say, &quot;The AC peak voltage must not exceed a couple of millivolts in peak magnitude?&quot; That's easy to read and directly addresses the situation and you'd still also know that the base-emitter junction is forward-biased, too.
I don't know why they didn't just do that. You'd then know that you aren't free to make much adjustment to the AC voltage and things would be a lot better for the example at hand. (It would keep you nearer the much simpler &quot;linearized&quot; small-signal model of the BJT and well away from its global, highly non-linear behavior.)