This gets pretty messy (though it’s always possible that I’ve simply overlooked a nicer approach).
Number the people $1$ through $25$. To form one set of partnerships you can first choose which of the $25$ will be unpartnered. Then repeatedly assign a partner to the lowest-numbered person remaining; this can be done in $23\cdot21\cdot19\cdot\ldots\cdot1=\frac{24!}{24\cdot22\cdot\ldots\cdot2}=\frac{24!}{2^{12}12!}$ ways, so there are $25\cdot23\cdot21\cdot19\cdot\ldots\cdot1=\frac{25!}{2^{12}12!}=7,905,853,580,625$ possible pairings.
There is no harm in assuming that the original pairing leaves $25$ unpartnered. Say that a pairing is good if it shares exactly $8$ couples with the original pairing. We’ll count the good pairings in two groups.
Consider first a good pairing that leaves $25$ unpartnered. There are $\binom{12}8$ ways to choose which $8$ couples it shares with $P$. The other $4$ couples must all be broken up. Say that the couples involved are $\{p_1,p_2\},\{p_3,p_4\},\{p_5,p_6\}$, and $\{p_7,p_8\}$. Now $p_1$ and $p_2$ can get new partners from the same original couple or from different original couples.
- If they come from the same original couple, there are $3$ ways to choose that couple and then $2$ ways to make the new pairings, for a total of $6$ ways to form the new couples. The remaining two original couples can then reform in $2$ ways, so there are altogether $3\cdot2\cdot2=12$ such ways to break up the $4$ couples.
- There are $6\cdot4$ ways for $p_1$ and $p_2$ to get new partners from different original couples. The remaining $4$ people consist of one original couple and two other people; the original couple has to be broken up, so these $4$ can be paired up in $2$ ways. Thus, there are $6\cdot4\cdot2=48$ such ways to break up the $4$ couples.
There are therefore $\binom{12}8(12+48)=60\binom{12}8=29,700$ good pairings with $25$ unpartnered.
Now consider a good pairing that leaves some $k\ne25$ unpartnered. We retain $8$ of the other $11$ couples. If $25$ ends up with $k$’s original partner, that leaves $3$ original couples to be broken up, say $\{p_1,p_2\},\{p_3,p_4\}$, and $\{p_5,p_6\}$. If $25$ ends up with someone else, it leaves $2$ original couples and two ‘strays’, say $\{p_1,p_2\},\{p_3,p_4\},p_5$, and $p_6$.
- In the first case there are $24$ choices for $k$ and $\binom{11}8$ choices for the couples to be retained. Then $p_1$ and $p_2$ must get new partners from different original couples, which can happen in $4\cdot2$ ways, so there are $8\cdot24\cdot\binom{11}8$ good pairings of this type.
- In the second case there are $24$ choices for $k$, $22$ choices for $25$’s partner, and again $\binom{11}8$ choices for the couples to be retained. Then either $p_5$ and $p_6$ are partnered, in which case there are just $2$ ways to repartner $p_1,p_2,p_3$, and $p_4$; or $p_5$ and $p_6$ are partnered with members of different original couples, which can be done in $4\cdot2=8$ ways and allows no further choice. There are therefore $10\cdot24\cdot22\cdot\binom{11}8$ good pairings of this type.
This gives us a total of $228\cdot24\cdot\binom{11}8=902,880$ good pairings with $25$ partnered, and a grand total of $932,580$ good pairings.
The probability of getting a good pairing is $\frac{932,580}{7,905,853,580,625}\approx 1.1796\times 10^{-7}\;.$