PNP too big of a voltage drop accross Collector Emitter
I suppose the problem is that during start-up you are not providing enough base current to \$T_1\$ because of the capacitor \$C_3\$. Once \$V_{CTRL}\$ is pulled high, you are slowly charging the base capacitor \$C_3\$ of \$T_2\$ before turning it on. Reducing \$C_3\$ and \$R_6\$ should turn on the logic / power BJT faster, thus causing a smaller voltage drop across the power PNP transistor.

Edit #1

Considering the current gain for the case where the collector current \$I_C\approx -3A\$, you would need a base resistance smaller than:

$$R_4&lt; \dfrac{12V - V_{BE, T1} - V_{CE, T2}}{I_B}$$

$$R_4&lt; \dfrac{12V - 1.1V - 300mV}{\dfrac{3A}{75}} \approx 265\Omega$$

Where \$V_{CE, T2}\$ is assumed to be the approximate collector emitter voltage of a common logic BJT.