In your degenerated diagram, the left side PNP is diode-connected. So to a first order, you can treat it as a diode in series with the two resistors of \$1\:\text{k}\Omega\$ and \$18\:\text{k}\Omega\$. Without doing the calculations, this should be approximately \$1\:\text{mA}\$. That current will be present throughout the series group, so \$1\:\text{k}\Omega\$ times \$1\:\text{mA}\$ gives about \$1\:\text{V}\$ across the emitter resistor. So the base will be about a diode drop lower, or around \$18.3\:\text{V}\$. This will be mirrored over to the right side PNP and so its collector current should be (if connected to a load that doesn't force saturation) about \$1\:\text{mA}\$, as well.
Assuming both BJTs are thermally coupled, the degeneration buys some immunity to part variations.
In the case of the current mirror without degeneration, variations in saturation current and emission coefficients lead to variations in base-emitter voltage vs collector current. And naked (no degeneration), there's a 10-fold increase in collector current for a \$60\:\text{mV}\$ variance in base-emitter voltage. Since two different parts might easily have \$30\:\text{mV}\$ variance between them, unmatched, this can mean a factor of 3 difference in collector currents (either way.) So that's terrible.
With the degeneration added, a \$30\:\text{mV}\$ variance means \$\approx 30\:\mu\text{A}\$ difference, which is much better in response to BJT variations alone (temperatures are assumed the same.)
A price is the voltage overhead.
More technically, though, you cannot assume a standard base-emitter (diode) drop for the left side PNP. It's actually going to behave according to the highly non-linear Shockley equation. So if you want a precise equation it takes some chain of mathematics to produce the LambertW equation that does make this calculation without iterating over and over. And you need to take into account base recombination currents, as well, while doing so. And BJTs also have an Early Effect that comes in with a 2nd order effect that especially applies to the right side PNP depending on its load. It's possible to account for all of this and develop a formula for both BJTs. And you can analyze that formula using sensitivity equations for whatever parameter you want, then. But this is usually reserved for people writing short "white papers."
A still more interesting question is how this all behaves if there is degeneration on only one side and not the other or where the degeneration resistor values are significantly different from each other. The answers are no longer quite so simple.