The following is merely to close with a simplified ‘description’ of the one solution produced here as a result of initially committing to the idea that the Z32 has been given to us as a problem to be solved. To aid succinctness, this summary description is intentionally devoid of any discussion as to what may have led to any one of the steps given. Likewise, it is presented in that way which gets to the result quickest, without consideration as to whether this reflects a plausible ordering of steps during any reasonable attack on solving the problem.

The proposed solution, then:

1. In the Z13, have the three ‘8-ball’ symbols encode ‘I’. (View.)
2. Likewise, in the Z32, find the three symbols that are present in the Z408 and there encode ‘I’ also, and give these the same encoding here, including where duplicated. (View.)
3. Interpret, from the numbering around the compass rose on the Phillips map, the expected unit for the bearing in the Z32 as ‘SIGNS’ and fit this word around the ‘I’ as given in step 2. with the encoding reverse-‘K’. Duplicate the first ‘S’ (plaintext) under the cipher encoding of ‘O’ to where this is repeated later in the cipher. (View.)
4. Complete ‘EIGHT’ at the very beginning of the cipher and transfer the resultant ‘E’ (plaintext) to its equivalent place later in the cipher, as given by the duplication on the very first symbol. (View.)
5. Directly transfer the encodings obtained thus far for the symbols that are common to both the Z32 and the Z13 across to the latter and match all duplicates there. ‘SIGNS’ is completed in the Z13 to evidence the presence, also, of one position-transposed character, contextually having been shifted from an original position at 2 in the cipher. (View.)
6. Interpret the last sequence of 6 characters in the Z32 as being equivalent, as to plaintext content, to the span of equal length from characters 2–7 in the Z13. Note that this uniquely associates the three ‘8-balls’ in the Z13 encoding ‘I’ with the three symbols in the Z32 used in the Z408 to encode ‘I’, supporting the choices made at the start in 1. and 2. (View.)
7. Interpret the two-character sequence prior to this last in the Z32 as being likewise equivalent, as to plaintext content, to the remaining unaccounted-for characters at 1 and 11 in the Z13. (View.)
8. From 7., complete the cross-encodings of each of these two characters where it is absent in the one or the other cipher. This gives ‘E’ (plaintext) for the second letter of the remaining 6-letter word on the duplicate ‘N’ (cipher text) in the Z13. (View.)
9. Identify that this latter can only be reasonably completed with either ‘GEMINI’ or ‘MEDICI’, with a heavy favour towards the former on contextual grounds. (View.)
10. Complete the remaining block of the Z32 cipher by inserting ‘PLUSFOURINCHES’, as the only one of two alternatives which makes landfall on the Phillips map. (View.)
11. Interpret the thrice-repeated ‘8-ball’ symbol in the Z13 — the only inherently numerical symbol occurring anywhere in the ciphers as a whole — as intended to indicate the value of the bearing (‘8 signs’) in relation to this cipher.

As stated, this would not be the route actually taken towards solution in practice. However I believe, in the manner of a ‘walkthrough’, it does capture the essence of the solution in a way which best facilitates welcome peer evaluation.

Try it! See what you think.

In conclusion, from the considered study and evaluation presented here, my own personal assessment as to the result would be:

… that, through the resulting co-solution of the Z13 and the Z32 to convey precisely the same information (minus the range, in its entirety, in the case of the shorter Z13), founded as this is upon a necessary cross-identification of the three ‘8-ball’ symbols in the Z13 uniquely with the occurrence in the Z32 of the three symbols encoding ‘I’ in the earlier Z408 (p = 0.004) — and which cross-identification unequivocally fixes the units for the bearing, the value for the bearing and the range, and further unambiguously completes the text of the Z13/32 beyond the statement of the bearing/range — the probability of this co-solution not being correct for both must be assessed as very, if not exceedingly, small.