Neither of these properties implies the other:
There is an admissible algebra that is not pseudo-compact
If $A$ is a ring with the discrete topology that is not artinian then it is admissible but not pseudocompact.
Less degenerately if $A$ is a noetherian ring and $I$ is an ideal such that $A/I$ is not artinian then the $I$-adic completion of $A$ is admissible but not pseudocompact --- the first example is simply the case $I=0$.
There is a pseudo-compact ring that is not admissible
If $G$ is any profinite group then the complete group algebra $K[[G]]$ over an algebraically closed field $K$ of characteristic zero, in the sense of https://www.ams.org/journals/bull/1966-72-02/S0002-9904-1966-11513-6/S0002-9904-1966-11513-6.pdf, is pseudocompact.
I guess you are only interested in commutative rings but if we take $G$ to be the profinite completion of the integers then $K[[G]]$ is commutative.
Now for every prime $p$, there is a quotient $C_p$ of $G$ that is cyclic of order $p$. Moreover there is non-trivial homomorphism $\theta_p\colon K[[G]]\to K$ that factors through $C_p$. For each $p$, $\ker \theta_p$ is an open (maximal) ideal. Moreover if $p\neq q$ then $\ker \theta_p\neq \ker \theta_q$.
If an open ideal $I$ has the property that for every $p$ there is some $n$ such that $I^n\subset \ker \theta_p$, then since each $\ker \theta_p$ is maximal, $I$ is contained in the intersection of all the $\ker \theta_p$. It follows that $K[[G]]/I$ is not artinian since it has infinitely many maximal ideals.
Thus $K[[G]]$ is pseudo-compact but not admissible.