What exactly does a Lorentz transformation provide? 
The primed reference frame is moving relative to the unprimed frame. So if we were to take the lorentz transformation of point P from the unprimed to primed, would it be the point A or B that it returns ? Assuming that the Lorentz transformation is passive i.e. we are talking about the same event in two frames, my first guess was B, because only at B will the primed reference frame see P to occur. A on the other hand is what the unprimed reference frame sees when P occurs. Is this true ?
PA is parallel to the x axis and PB is parallel to the x' axis.
 A: See below for an update to address the "passive transformation".

The [active] Lorentz-boost-transformation of event P is neither your events A nor B,
but an event between your A and B
that is on the hyperbola centered at the origin that passes through P.
The boosted event is joined to the original event by a line
that is not parallel to either axis of these two frames.
From my spacetime diagrammer https://www.desmos.com/calculator/emqe6uyzha ,

the boosted events lie on a common hyperbola (which are events that "equidistant in future time from the meeting event" according to the Minkowski metric).
The spatial-axis of a frame is parallel to the tangent to that hyperbola (shown as a dotted line).
It may be instructive to consider the Euclidean version of the construction (move the slider to $E=-1$):

The rotated point is joined to the original point by a line
that is not parallel to either axis of these two coordinate systems.

UPDATE:
Concerning the passive transformation...
the events are left unchanged in spacetime.
But now the passive Lorentz boost provides the coordinates
of an event in another inertial frame.
In this case, your event B is the event simultaneous with P, according to the moving inertial observer.
The diagram below is for $v=(3/5)c$.
So, the transformation would assign the event $P$ with red-coordinates $(t=1,x=0)$
with blue coordinates $(t=1.25, x=-0.75)$.
The passive transformation identifies event B as the event the moving inertial observer says is simultaneous with P
and calculates the blue time-coordinate of P to be equal to the blue time-coordinate of B as $t=1.25$.
(The transformation will also identify an event (not shown in the image) on the moving x-axis (at blue time t=0) that is "at the same place as P" according to blue. The spatial-coordinate of that event is $x=-0.75$.)

A: 
Look at this Minkowski diagram point P move parallel to the x‘ axis , the event point is then $~E(ct‘,x‘)$
