Order 3 projective plane with playing cards The projective space based on $\mathbb{F}_3$, the finite field with 3 elements (PP3), has 13 lines and 13 points, with every two lines meeting at one point and every two points determining a line. Each of the 13 lines consists of four points. A possible listing would be A23T, 456T, 789T,  A47J, 258J, 369J, A59Q, 267Q, 348Q, A68K, 249K, 357K, and TJQK. I note that 13 lines with 4 points to a line suggests using playing cards, explaining my notation, and indeed I got out a set of playing cards and dealt them out into a 4x13 rectangle that represents PP3. But playing cards have suits too, and I rearranged the cards so that each line contains one card from each suit. The resulting arrangement seemed haphazard, so I tried it with the Fano plane, using 3 suits and 7 lines. I found that the 124, 235, 346, 457, 561, 612, 723 arrangement (1 = Ace) can be obtained by adding 1 mod 7 to each card in a line to get the next line. In that case, then let each first element of the line be a heart, each second a diamond, and each third a spade.  I tried doing this with PP3 and it didn’t seem to work. Is there such an arrangement for PP3, where the points represent elements of $\mathbb{Z}_{13}$? Is there an orderly way of assigning suits to the cards? And how many ways are there of assigning suits to the points in each line in PP3?
 A: Short answers

Is there such an arrangement for PP3, where the points represent elements of $\mathbb Z_{13}$?

Using $(0, 1, 3, 9)$ as the relative numeric values of the points on one of the lines yields the property you requested.
Of course, you can also re-order the points, scale or negate offsets (mod 13), or use another element as zero or maybe not even have a zero offset. So depending on how you count things, you might find a large collection of viable offsets, but they could all be inferred from this one. Among these alternatives, $(0,1,4,6)$, $(0,2,5,6)$ and $(0,2,3,7)$ would each lead to other card values being considered as collinear, so in that sense you might say there are four possible arrangements to consider.
The $(0,1,3,9)$ configuration could be laid out like this, with one suite per row:
A 2 3 4 5 6 7 8 9 T J Q K  ♢
2 3 4 5 6 7 8 9 T J Q K A  ♡
4 5 6 7 8 9 T J Q K A 2 3  ♠
T J Q K A 2 3 4 5 6 7 8 9  ♣

I have verified that this indeed has the structure of PP3. A previous version of this answer had stated that no such solution exists, but the code leading to that had a mistake.

And how many ways are there of assigning suits to the points in each line in PP3?

There are $80{,}640\cdot 4!=1{,}935{,}360$ possible assignments of card suites. $352$ of these assignments appear to be sufficiently distinct to be interesting, and I have a list of them.
$8$ of these are particularly symmetric: they have a 13-fold symmetry, i.e. remain invariant under 13 projective transformations. For these one can find a $\mathbb Z_{13}$ description.

Is there an orderly way of assigning suits to the cards?

I think I have what you were looking for. For other readers with different requirements, perhaps my list can serve as a basis for further exploration.
How to find out
I've written a bit of code, trying to answer your questions. I'll describe in more detail what I did, and also what I found.
In a first step to find viable assignments, I've modeled this as an exact set cover problem. Let $P$ be the set of all points, and $L$ the set of all lines. Let $I\subset P\times L$ be the set of incidences. And let $S$ be the set of suites. One way of formulating your question is to look for a function $I\to S$ which assigns a suite to each incidence in such a way that each line has exactly one incidence of each suite (because you want each suite to occur once per column in your arrangement of cards), and also each point has exactly one incidence of each suite (because that's how the deck of cards is composed, each value occurs once per suite).
I'm thinking of the set cover problem as a big binary matrix, with the rows as sets and the columns as set elements. The rows would correspond to $I\times S$, i.e. each row assigns one suite to one incidence. I have three groups of columns. One group corresponds to the elements of $I$. Selecting exactly one row for each incidence ensures that there is exactly one suite for each, making $I\to S$ an actual function. The second group of columns corresponds to $L\times S$. Selecting exactly one row for each line-suite combination ensures that the four incidences of each line correspond to four different suites. The third group corresponds to $P\times S$, ensuring different suites for the incidences of a point. Each row has three ones, one in each group, in the places corresponding to the relevant pieces of the row information.
So now the problem of finding a suite assignment matching your requirements is equivalent to selecting some of the rows so that they cover all of the columns, i.e. so that each column has a one in exactly one of the selected rows. I happened to have an implementation of DLX lying around, so I fed the matrix I just described to that. As a slight simplification, I forced the first four incidences to correspond to the four suites in order, eliminating a factor $4!=24$ that would result from permutating the suites. I obtained $80{,}640$ exact set covers from this.
Next I eliminated solutions that could be obtained from other solutions using projective transformations. I enumerated all projective transformations that map any of the 13 lines to the first line, with the points in order of ascending suite. If any of them resulted in a solution that was lexicographically smaller I dropped it. This left me with $352$ solutions which are unique under projective transformations. I'm sharing the list of these solutions.
I also used these projective transformations to characterize symmetries: I counted how many different results I obtained using different transformations. For most solutions I got $13\cdot3\cdot3\cdot2=234$ different descriptions, one for every projective transformation I tried. But for $8$ of the resulting $352$ solutions I only got $3\cdot3\cdot2=18$ different descriptions. This means that each distinct description had $13$ projective transformations leading to them. So each of these had a $13$-fold projective symmetry under which it remained unchanged.
Now to look for the $\mathbb Z_{13}$ structure. I imported my solutions into Sage, and turned each solution into a $(13\cdot 3+1)\times(13+3)$ matrix over $\mathbb Z_{13}$, a system of linear equations. The first 13 columns (and the corresponding unknowns) represent the numeric value of each point of the plane, the last three the difference in value from one suite to the next. Row $3i+j$ expresses that the difference for line $i$ of the plane between the point of suite $j$ and the point of suite $j+1$ must match the difference associated with these suits. The last row of the matrix just has a one in the first column. It forces the first card value to be zero, which avoids a cyclic rotation of card values to show up in the solution.
Any valid $\mathbb Z_{13}$ structure would need to show up as a non-zero solution of this homogeneous set of equations. So I computed the dimension of the right kernel for the matrix in each of the 352 different cases I had. (In a first version of this answer I had a mistake here, and computed the kernel for a single solution 352 times). For the 8 symmetric arrangements, this led to a one-dimensional kernel, i.e. a space of point values and offsets which would satisfy all requirements. A quck inspection confirmed that the point values were all different. From the offset values in these solutions I obtained the $(0,1,3,9)$ offset vector mentioned at the top of my post. I also got $(0,1,4,6)$ but later found that the two are essentially the same once you take multiplication of offsets with an element of $\mathbb Z_{13}$ into account.
