Will the puck rotate without slipping in this situation? 
We have two pucks moving on a plane without friction. On one of them a force is applied on it's center of mass. On the second a force of equal magnitude is acting tangential to the puck and at a distance equal to it's radius.

We can prove that the acceleration of both the pucks would be same, and so the CM of both would move together (if the forces are applied at the same time).
For the first puck , $$F_T. \triangle X_{CM} =\frac{1}{2}mv^2 \tag{1} $$
For the second puck , the hand has to travel a longer distance, so the force is applied over a distance $\triangle X_{CM} + d$. Hence, $$F_T.(\triangle X_{CM} + d)=\frac{1}{2}mv^2+\frac{1}{2}I \omega^2 \tag{2} $$
Solving equation (1) and (2) , we get the rotational work done, $$F_T.d=\frac{1}{2}I \omega^2$$
So, $$\vec{F_T}.\vec{d}=(\vec{R} \times \vec{F_T}).\vec{\theta} \\ \ \ \ \ \ \ \ \ \  = \vec{F_T}.(\vec{\theta} \times \vec{R})$$
Hence, $$\vec{d}=\vec{\theta} \times \vec{R}$$
This would imply that the puck rotated without slipping, because the length of string unwound is the total arc distance rotated.

But, $$\vec{V_Q}=\vec{V_{CM}}+\vec{\omega} \times \vec{R}$$ and $$\vec{V_P}=\vec{V_{CM}}$$ Hence there is relative motion between $V_Q$ and $V_P$, and hence slippage occurs.
So does the puck rotate without slipping or does it not? How can slippage not occur if there is relative motion between point $P$ and $Q$?
 A: There is no contradiction.
$v_Q$ (the instantaneous horizontal peripherical speed) and $v_P$ (the horizontal speed of the string) are the same when $Q$ is at the top. But around every turn $v_Q$ becomes smaller and reach a minimum when $Q$ is at the bottom.
If there is no slippage, $v_Q = v_P = 2v_{CM}$ when at the top, and $v_Q = 0$ at the bottom.
In this case, $v_{CM} = \omega R$, and $d = \Delta X_{CM}$ on $(2)$.
By the way, the equation $(2)$ is only valid for the no-slip case. We can say for an arbitrary friction force $f_{fr}$:
$$F_T-f_{fr} = m\frac{dv_{CM}}{dt}$$
$$\tau = (F_T+f_{fr})R = I\frac{d\omega}{dt} \implies F_T+f_{fr} = \frac{I}{R}\frac{d\omega}{dt} $$
Adding the equations:
$$F_T = \frac{1}{2}m\frac{dv_{CM}}{dt} + \frac{1}{2}\frac{I}{R}\frac{d\omega}{dt}$$
In the case of no-slip, the displacement of $F_T$ is twice that of the CM, and $v_{CM} = \omega R$
$$F_T.(2dX_{CM}) = m\frac{dv_{CM}}{dt}(dX_{CM}) + \frac{I}{R}\frac{d\omega}{dt}(dX_{CM})$$
$$F_T.(2dX_{CM}) = mv_{CM}dv_{CM} + \frac{I}{R}d\omega (\omega R)$$
$$dw = F_T.(dX_{CM} + \delta) = d\left(\frac{1}{2}mv_{CM}^2\right) + d\left(\frac{1}{2}I\omega^2\right)$$
where $\delta = dX_{CM}$
A: The problem here is the assumption that $V_P=V_{CM}$
Your first thought that the motion of the centres of mass will be identical for the same total amplitude and direction of external force is correct, if the impulse (force*time) is the same, and both pucks will slide at the same speed.
The energy consideration works for the rotational speed as well. Point $P$ on the string is moving faster than $V_CM$ in the second case, as the distance travelled by the string is longer by $d$, so $V_P=V_Q=V_{CM}+\omega\times R$.
The total length the string travels is $\Delta x+d$, not $d$ - the puck has slid a distance of $\Delta x$ and unwound an additional $d$ of string.
