What an odd question. Let $K$ be the foot of the perpendicular from $D$ onto $AB$. Let $L$ and $M$ be defined such that $KDLM$ is a square, and $M$ lies on $AB$. Suppose $AB$ and $CD$ not parallel. Extend $CL$ to meet $AB$ at $X$. Now scale $KDLM$ towards $X$ with some factor such that the points don't coincide with any of $A, B, C, D$. Let $K'$ be $A'$, $M'$ be $B'$, $L'$ be $C'$, and $D'$ remains what it is. This square fulfils the question requirements, with the lines meeting at $X$. If $AB$ and $CD$ are parallel, but $BC$ and $AD$ aren't, then we can do a similar strategy, just permuting the letter labels. If $ABCD$ is a rectangle, WLOG $AB\ge BC$. Construct a square $DBC'A'$, then project $C'C$ and $A'A$ to meet at $X$, then dilate the square about $X$ as above. $C'C$ and $A'A$ can't be parallel as (with $AB$ oriented above $CD$, and running horizontally) $C'C$ is to the left of $BC'$, while $A'A$ is to the right of $A'D$.

Very interesting question... If $AC \perp BD$, take $P \equiv AC \cap BD$, and it is trivial to construct a square $A'B'C'D'$ with center $P$ such that $AA', BB', CC', DD'$ are concurrent at $P.$ Thus, suppose that $AC \not\perp BD.$ We will construct a point $P \in BD$ distinct from $B, D$ such that $\angle APB = \angle CPB.$ Let the perpendicular bisector of $\overline{AC}$ meet $BD$ at $Q$ and let $P$ be the second intersection of $BD$ with $\odot(ACQ).$ Note that $Q$ is a midpoint of arc $\widehat{AC}$ on $\odot(ACQ)$ because $QA = QC.$ Therefore, $BD$ bisects $\angle APC.$ If $P$ coincides with $B$, repeat above construction about vertex $D$ instead of vertex $B.$ If $P$ then coincides with $D$, we have $\angle ABD = \angle CBD$ and $\angle ADB = \angle ACB.$ Therefore, $ABCD$ is a kite, implying that $AC \perp BD$, impossible. Thus, we may assume assume that $P \not\equiv B.$ Now, choose $A_1, C_1$ on $PA, PC$ such that $\angle A_1BP = \angle C_1BP = 45^{\circ}$ and let $D_1$ be the reflection of $B$ in $A_1C_1.$ Since $\angle A_1PB = \angle C_1PB$ it follows that $\triangle PBA_1 \cong \triangle PCA_1.$ Hence, $BA_1 = BC_1.$ Together with $\angle A_1BC_1 = 90^{\circ}$ it follows that $A_1BC_1D_1$ is a square. Moreover, $D_1 \in BD$ by symmetry. To finish, we dilate $A_1BC_1D_1$ with center $P$ to a new square $A'B'C'D'$ with all vertices distinct from $ABCD.$ Then $AA', BB', CC', DD'$ are concurrent at $P$, as desired. $\square$

Doesnt a projectivity help here? Specifically there exists a projective trabsformation which sends a one quadrilateral to another. as a special case a Quadrilateral to a square. since the projecticity is centered from a point the lines $AA',BB',CC',DD'$ are concurrent from th point of perspectivity. I dont get whats wrong with this.... It cant be this trivial

Technically that approach is invalid, as a projective transformation is not necessarily planar. However, that idea is correct, as the specific projective transformation mapping quadrilaterals is planar.

So essentially can this idea be ammended to solve the problem. Also why is a projective transformation not planar, and how does this make the solution invalid? Since we are in 2-dimensions.

If $AC\perp BD$ then we are done by taking a square centered at the intersection of the diagonal. Now suppose $AC$ is not perpendicular to $BD$. We call the vertex $D$ bad if $\angle ADB=\angle CDB=45^{\circ}$ and good otherwise, define good and bad similarly for $A,B,C$. Now notice that if at least two of the vertices are bad then $ABCD$ must be a square which contradicts the assumption that $AC$ is not perpendicular to $BD$. Therefore, at least one of the two sets $\{A,C\}$ and $\{B,D\}$, say $\{B,D\}$ consists only of good vertices. Now take $BD$ to be the $x-$axis and let $A=(a,b)$, $C=(c,d)$. For each $m\in \mathbb R,n\neq 0$ defined a $(m,n)-$square to be a square centered at a point $(m,0)$ and has vertices $V_A=(m,n),V_C=(m,-n),V_B=(m-n,0),V_D=(m+n,0)$ CLAIM. There exists a $(m,n)-$ square such that $V_A\neq A$,$V_B\neq B$, $V_C\neq C$, $V_D\neq D$ and that $AV_A$ and $CV_C$ are concurrent on the $x-$ axis. Proof. This is just coordinate bashing. Indeed, the $x-$ intercept of $AV_A$ and $CV_C$ are $$x=a-\frac{b(m-a)}{n-b}\text{ and }x=c+\frac{d(m-c)}{n+d}$$respectively. Therefore the concurrency is equivalent to \begin{align*} a-c&=\frac{d(m-c)}{n+d}+\frac{b(m-a)}{n-b}\\ \iff m(d+b)&=(a-c)n+ad+bc\\ \end{align*}Now if $d+b\neq 0$ or $ad+bc\neq 0$ then there exists infinitely many nonzero solution $(m,n)$. If both of them are zero then we must have $a=c$, this implies $AC\perp BD$, contradiction. If $\frac{d+b}{a-c}\neq \pm 1$ then we can find $m,n$ such that $(m+n,0),(m-n,0)$ are distinct from $B,D$. If it is $1$ then $(m-n,0)=(ad-bc,0)$, if this is the coordinate of $B$ then $B$ is bad, contradiction, the case when this fraction is $-1$ is similar. $\blacksquare$ This claim obviously completes the problem.

However,in Joyce Tan's proof,in the case that ABCD is a rectangle,the order of the verticals of the square will be A'C'B'D',which doesn't meet the requirements.