We can classify all $\binom 8 4$ functions of three variables with $4$ minterms according to the "number of blocks" one obtains on a Karnaugh map. First note that the K-map method produces a prime and irredundant expression: every term (or block) is a prime implicant, and every term covers one minterm not covered by any other term in the expression.
We classify prime and irredundant covers according to the number of prime implicants. Then we'll argue that the correspondence between covers and functions is one-to-one.
- Prime and irredundant covers with four implicants. There are two of them, even and odd parity. No two minterms may be adjacent on the map, and there's only two ways to achieve that.
- Prime and irredundant covers with three prime implicants. There are eight of them of the type shown by @DilipSarwate, namely $(x \wedge y) \vee (x \wedge z) \vee (y \wedge z)$. There are 24 more of the form $(x \wedge y \wedge z) \vee (\neg x \wedge \neg y) \vee (\neg x \wedge \neg z)$.
- Prime and irredundant covers with two prime implicants. We divide them according to the number of variables appearing in the cover.
- Two variables: these are the even and odd parity functions of two variables, of which there are six in total.
- Three variables: These covers must be of the form $(v \wedge b) \vee (\neg v \wedge c)$, where $v$ is a variable and $b$, $c$ are literals (not of $v$). The variable $v$ can be chosen in three ways; for each choice of $v$, there are four choices of $b$ and then two choices for $c$, for a total of 24 covers.
- Prime and irredundant covers with one prime implicant. The implicant is a literal; hence there are six choices in this case.
In summary: $2 + (8+24) + (6+24) + 6 = 70 = \binom 8 4$ as expected. The number of covers that consist of three "blocks of two" is 8, as identified by @DilipSarwate.
We now argue that the each of the 70 functions has exactly one prime and irredundant cover. This is not necessary if we are convinced we counted all the covers above, but it doesn't hurt either. Recall that the consensus of $a \wedge b$ and $\neg a \wedge c$ is $b \wedge c$. Further recall that a term of a prime cover is essential (i.e., must be part of any prime cover) if it is not covered by the disjunction of its conjunctions and consensus terms with the other prime implicants of the cover.
Finally, if a prime cover consists of all essential primes, it is the unique prime and irredundant cover. All these are classic results. Since all covers listed above are made of essential primes, they are all unique.