Qualitatively, yes, the second transistor will load the first. Qualitatively is not good enough, however. You have to run the numbers and see QUANTITATIVELY what is going on: how MUCH will Q2 load Q1?
Here's the quick-and-dirty sanity check analysis.
Vb1 = 1/3 Vcc = 5V
Divider current = Vcc / (Rba+Rbb) = 15/(10K+5K) = 1mA.
Ve1 = Vb1 - 0.7V = 4.3V
Ie1 = Ve1/Re1 = 1.4333... mA
Ic = Ie x beta/(beta + 1). Usually, this is approximated as Ic = Ie.
From the 2N3904 datasheet, at Ic = 1 mA, beta = 70.
Ib1 = Ic1 / Beta = 20.5 uA, which does not significantly disturb the voltage divider output.
Vc1 = Vcc - Ic1 x Rc1 = 15 - 1.4333 mA x 5K = 7.8333 V.
Vb2 = Vc1, because of circuit topology.
Ve2 = Vb2 + 0.7V (because it is a PNP transistor) = 8.5333 V
Ie2 = (Ve2 - Vcc) / Re2 = (8.5333 - 15) / 2K = -3.2 mA.
Ic2 = Ie2
From the 2N3906 datasheet, at Ic = -1.0 mA, beta = 80 (minimum).
Ib2 = Ic2 / Beta = -40 uA, which is negligible compared to 1.4333 mA.
So the load from the second transistor does not SIGNIFICANTLY affect the operating point of the first transistor.
The above numbers are not exact, but neither are the transistor datasheet beta values. Both datasheets give the stated minimum values, and state a maximum beta of 300. If the beta for Q2 is higher than the minimum value, then the load presented by Q2 to Q1 is lower, meaning even less disturbance than the already-negligible -40 uA.