Sampling theory terminology This is more of a terminology question, but if references can be given for more information that would also be great.
Assume instead of sampling actual people we are sampling from items that belong to them. If at least one item belonging to a person is in the sample that person is also in our sample. Given the total number of items is known, estimate the unknown number of people in the population.
What is the name for this type of problem? Imperfect sampling, partial sampling?
Simple example:


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*Alice has 5 books

*Bob has 2 books

*Charlie has 7 books

*Daisy has 1 book


We don't know that four people is the truth, but we do know 15 books are total. We sample in some fashion and see how many unique people those books belong to


*

*Alice has 1 book in sample

*Bob has 2 books in sample

*Charlie has 4 book in sample

*[Daisy in unknown in sample, as 0 books]


This yields 7 sampled books out of 15 with 3 unique people. First approximation would suggest simple proportions to solve (15/x)=(7/3) yielding an estimate of 6.4 unique people in population.
 A: Your presentation is a bit nonstandard application of the ratio estimator, which has a long tradition in sampling statistics. You have a population of $N$ books from which you take a simple sample without replacement (SRSWOR; I assume you can give a definition of an SRS as it is a bit nontrivial) of size $n$. You count a distinct number of people $k$ in the sample and produce the ratio of the inverse of number of books per person $r=k/n$. Then the estimated number of people in the population is $\hat K=N \cdot r = N k/n$.
To compute the standard error around that estimate, you need to define the book-level variate on the same scale as $r$, which will probably have to be the inverse number of books per person for the person that the book is identified with (i.e., $r_i=1$ for one Alice's book, $r_i=0.5$ for the two Bob's books, and $r_i=0.25$ for the four Charlie's books).
They all are, however, underestimates of how many books a person has. A better estimate would inflate these per-person estimates by the inverse of the sampling rates. If you sampled at the rate of $7/15$, you cannot realistically expect to get every book a person has, but you can expect to get $7/15$ of all of their books. Hence a better ratio estimate of the number of books per person is $\tilde r^{-1} = n/k \cdot N/n = N/k$. Then the estimate of the population total number of people is $\tilde K=N \cdot \tilde r=N \cdot k/N=k$, which is not satisfactory as it is clearly only a lower bound (there may be some people who did not make it to the sample, like Daisy).
If you were to know the true number of books per person (that Alice has 5, Bob has 2, and Charlie has 7),  then you could form your estimator of the population number of people as follows. Among the three sampled people, the inverse of the average number of books is $\breve r^{-1}=(5+2+7)/3=14/3$. The known people are known to have $\breve N=14$ books out of $N=15$ in the population, so there is $N-\breve N=1$ book(s) left. At the known rate of book ownership, these remaining book(s) can be estimated to contribute $(N-\breve N)r = 1\cdot 3/14=3/14$ persons, for a total of $k+(N-\breve N)r=3 3/14$ persons.
There is a further room for improvement on this, of course (provided that you know the true number of books). Take the remainder of the population of books $N-\breve N$, and entertain the possible ways of allocating it into persons, i.e., one-book persons, two-book persons, etc. Then you can maximize the joint likelihood of having observed the people that are in the sample and those who are not in the sample, and obtain the most accurate estimate of the number of people in the population. In the above example, you would trivially conclude that there must be exactly one person left out of the sample, as there is only one book left unaccounted for. If you had two books unaccounted for, you could decide to either split them to two one-book persons, or keep them together as a one two-book person; putting these into the multinomial probability calculation would tell you which of these two alternatives is more likely. If you had three books unaccounted for, you could have one person with three books, two persons with 1 and 2 books, or three persons one book each. Etc. For a person $j$ with $b_j$ books, the probability of not making into the sample is ${b_j \choose 0} {N-b_j \choose n}/{N \choose n}$, and the probability of being in the sample is the complement of that. The sample likelihood is then the complement of the probability of not being in the sample for those who made it to the sample (for Alice, Bob and Charlie: $1-{b_A + b_B + b_C \choose 0}{\sum b_j\choose n}/{N \choose n}$ times the probabilities for those who did not make it, ${\sum_j b_j \choose 0}{b_A + b_B + b_C \choose n}/{N \choose n}$, and this is maximized over the arrangements such that $\sum_j b_j + b_A + b_B + b_C = N, b_j \in \mathbf{N}$. This is basically saying that if you have a lot of books, you are more likely to make it to the sample. You need a discrete maximization algorithm (branch and bound?) to solve this. Once again, this relies heavily on your ability to figure out from a book how many other books that its owner have;  this may be a tall order depending on your actual application.
So this is not a terminology question at all, this is a rather serious methodological question.
