If $Y\sim\mu$ with probability $p$ and $Y\sim\kappa(X,\;\cdot\;)$ otherwise, what's the conditional distribution of $Y$ given $X$?

Some notation. When $\nu$ is a probability measure on a space $E$ and $\kappa$ is a Markov kernel on the same space, the semidirect product $\nu\rtimes \kappa$ is the measure on $E\times E$ (equipped with product $\sigma$-algebra) satisfying $$ (\nu\rtimes \kappa)(A\times B)=\nu(1_A\cdot \kappa 1_B). $$ It is the law of the first two steps of a Markov chain with initial distribution $\mu$ and transition kernel $\kappa$.

Formalizing the question. Let Ber$_p$ denote the probability measure on $\{0,1\}$ satisfying Ber$_p(\{1\})=p$. Consider the enlarged sample space $\Gamma=E^3\times \{0,1\}$ with the product $\sigma$-algebra, and equip $\Gamma$ with the probability measure $\mathbb P=\mu\otimes(\nu\rtimes \kappa)\otimes \textrm{Ber}_p$, where $\nu$ denotes the law of $X$.

Consider the function $f\colon \Gamma\to E$ given by $$ f(w,x,y,z)=\begin{cases}y,& z = 0\\ w,& z = 1\end{cases}. $$ When $f$ is regarded as a random element of $E$, it is precisely the result of "sampling from $\mu$ with probability $p$ and from $\kappa(X,\cdot)$ with probability $1-p$" in the way you have described.

Phrased in this precise and rigorous way, your question asks the following.

Reformulated question. For any $B\in\mathcal E$, determine the conditional probability $\mathbb P(f\in B\mid x)$.

You have guessed a formula for this conditional probability, which we will now verify.

Claim. The random variable $(1-p)\kappa(x, B)+p\mu(B)$ on $\Gamma$ is a version of $\mathbb P(f\in B\mid x)$.

In the proof of this claim, we will use notation like $\mathbb E[\textrm{variable};\textrm{conditions}]$ as a shorthand for the expectation of (variable times the indicator of the conditions) with respect to $\mathbb P$.

Proof. Unwinding the definition of conditional probability, the claim amounts to showing that $$ \mathbb P(f\in B,x\in A)=(1-p)\mathbb E[\kappa(x, B);x\in A]+p\mu(B)\mathbb P(x\in A)\tag{1}, $$ for all sets $A\in \mathcal E$. Splitting up the left side, we see that $$ \mathbb P(f\in B,x\in A)=\mathbb P(f\in B,z=0,x\in A)+\mathbb P(f\in B,z=1,x\in A). $$ On $z=0$, we have $f=y$ and on $z=1$, we have $f=w$. Thus $$ \mathbb P(f\in B,x\in A)=\mathbb P(y\in B,z=0,x\in A)+\mathbb P(w\in B,z=1,x\in A). $$ Using independence (coming from the product structure of $\mathbb P$) then yields $$ \mathbb P(f\in B,x\in A)=(1-p)\mathbb P(y\in B,x\in A)+p\mu(B)\mathbb P(x\in A). $$ Recalling that the law of $(x,y)$ is $\nu\rtimes \kappa$ and directly applying the definition of the semidirect product yields $\mathbb P(y\in B,x\in A)=\mathbb E[\kappa(x,B);x\in A]$. Substituting this into the previous display yields $(1)$, establishing the claim.