The load is to be placed at \$v_o\$.
The PNP is a current source, so two things are true:
1. It supplies bias, in place of the top resistor in your second diagram.
2. Because the diodes have a very low impedance when forward biased (the dynamic resistance of a diode goes as \$\frac{V_{TH}}{I_f}\$ (where Vth ~= 26mV), they act as a supernode (not quite, but close). That is: any change in current (AC), connected to any terminal of any diode, will have a very similar effect on the circuit.
To illustrate this more clearly, leave off the transistors and concentrate on the bias network (PNP or top resistor, two diodes, and R3 or bottom resistor). (This is reasonable because base current is small compared to the current flowing in this path.) The dynamic resistance of the diodes might be 10 ohms each, while the resistors might be 1k or more. The AC voltage drop across the diodes, in response to an AC input current on any node, is small, no matter which node it's applied to.
So you could very well use the second circuit, with the PNP connected to the tap, as shown. But this has an odd effect:
The PNP carries DC bias current, so the diode currents will not be equal. This isn't a problem as long as the difference is small -- but then your output voltage range will be small, because the PNP can only pull up so far over that range. (Think about the V = I*R change in the resistors.)
So the top resistor just makes things worse. Why not remove it, right?
That doubles the Thevenin resistance loading down the node -- that is, for a given input current \$\Delta i\$, you get twice the \$\Delta v\$. That's a win. But then the top diode isn't biased. So, let's just move the PNP up there, so its bias current goes in, and... et voila, we have the top circuit!
Tim