Understanding the role of a pull-up resistor
Drawn better, it looks like this:



simulate this circuit &ndash; Schematic created using CircuitLab
Within the left-box, you have a simple BJT used as a saturated switch. And on the right-box you have yet another such BJT used as a saturated switch. The trick, here, is that these are connected only via a capacitor!
With \$SW_3\$ is normally grounded, \$Q_4\$ is normally off and before \$SW_3\$ is engaged, the circuit powers up and \$R_{11}\$ and \$R_4\$ both &quot;pull up&quot; on each side of capacitor \$C_2\$, hauling both sides to \$+5\:\text{V}\$. But this is the same thing as saying that there is \$0\:\text{V}\$ across capacitor \$C_2\$. Yes? So, by default, there's no voltage across \$C_2\$.
Also, \$R_4\$ is supplying base current to \$Q_3\$. So \$Q_3\$ is on, by default. This means that OUT is close to ground, or near \$0\:\text{V}\$, by default.
When \$SW_3\$ is changed to the \$+3\:\text{V}\$ rail, it activates \$Q_4\$ causing it to suddenly pull down on its collector and therefore ground A. However, right at this moment in time please don't forget that there is also \$0\:\text{V}\$ across \$C_2\$. (See above paragraph.) So, it follows that B will equal A right at this moment. (This will change. But at the immediate instant the switch is changed, this is true.) This causes \$Q_3\$ to be turned off. And when that happens, \$R_5\$ is able to immediately pull OUT to \$+5\:\text{V}\$.
Now, this is what happens immediately. But if you hold \$SW_3\$ to this new state, then \$R_4\$ will start supplying current (as it now has \$+5\:\text{V}\$ across it.) This current will, at first, just start charging up the right-end of \$C_2\$ (the left end is held close to ground by \$SW_3\$ in its new state.) So as \$C_2\$ charges upward, the base voltage for \$Q_3\$ also starts rising upward. At some point, \$C_2\$ is sufficiently charged to have enough voltage across it that \$Q_3\$ is re-activated and pulls down on its collector causing OUT to return back near to \$0\:\text{V}\$.
If you keep \$SW_3\$ held, OUT will fully return to its \$0\:\text{V}\$ output. So from the perspective of OUT, you will see a &quot;rising towards \$+5\:\text{V}\$, but short-duration, pulse.&quot; Once \$C_2\$ charges up enough, \$Q_3\$'s base pulls away most of the current in \$R_4\$ and the charging of \$C_2\$ halts. And there it sits until you release \$SW_3\$.
Once you release \$SW_3\$, \$Q_4\$ turns back off and then \$R_{11}\$ can now pull upward on the left end of \$C_2\$. Since \$C_2\$ will only have about \$+700\:\text{mV}\$ across it (the right end is more positive than the left), the base of \$Q_3\$ will momentarily rise to \$+5.7\:\text{V}\$. \$Q_3\$ is already on and this voltage is way, way too much for the base-emitter junction of \$Q_3\$, so there will be a momentary high current pulse into the base of \$Q_3\$ as well as into the \$+5\:\text{V}\$ supply rail. But capacitor \$C_2\$ will also be rapidly discharged and will return to a state with \$0\:\text{V}\$ across it. The circuit is now back to where it was before.
The time period for the high-going pulse width would normally be proportional to \$\left(R_4+R_{11}\right)\cdot C_2\$. And in your circuit, this is on the order of about \$110\:\mu\text{s}\$. But it will likely be less than that. There is a significant added effect of the base-emitter junction of \$Q_3\$ and the resulting partial charged voltage involved. So it will always be less than that. I'll avoid worrying about the math, for now.