Let me describe a construction that meets your conditions. For this, let $(P,\succeq)$ be a poset (and always assume that $a\preceq b \Leftrightarrow b\succeq a$).

Stage 1. Let $(C,\succeq)$ be the Dedekind-MacNeille completion of $(P,\succeq)$, and let $h\colon P\to C\colon p\mapsto (p]$ be the order-embedding that takes an element $p\in P$ to its lower cone $(p] = \{x\in P\;|\;x\preceq p\}$.

Stage 2. Let $(D,\succeq)$ be the meet-subsemilattice $(C,\succeq)$ that is generated by the image $h(P)$. If this meet-semilattice has a least element, then label it $0$. If it does not have a least element, adjoin one and label it $0$.

At this point we have embedded $(P,\succeq)$ into a small meet-semilattice $(D,\succeq)$ with a zero element such that the maximal elements of $(P,\succeq)$ are mapped to the maximal elements of $(D,\succeq)$.

Stage 3. To each element of $d\in D$ that is not $0$ or an atom of $(D,\succeq)$ we shall adjoin a companion element $x_d$. Each new element $x_d$ will lie strictly below the element $d$ and also below any element $d'\in D$ for which $d'\succeq d$, $x_d$ will lie strictly above $0$, and $x_d$ will be incomparable with all other elements of $(D,\succeq)$. Let $(X,\succeq)$ be the poset obtained from $(D,\succeq)$ by adding all companion elements.

Stage 1 embeds $(P,\succeq)$ into a small complete lattice. Stage 2 ensures that Property 2 of the problem statement is satisfied. Stage 2 also ensures that maximal elements are preserved under the embedding. Stage 3 ensures that Property 1 of the problem statement is satisfied.

If you apply this construction to a chain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $2n-2$, which is much smaller than the size of the Boolean envelope of $P$. If you apply this construction to an antichain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $n+1$, which is also much smaller than the size of the Boolean envelope of $P$.

If $(P,\succeq)$ is a finite poset of order dimension $d$, then it can be shown that the order dimension of the poset $(X,\succeq)$ constructed above is at most $2d$. (To prove this for yourself, use the fact that the Dedekind-MacNeille completion does not increase order dimension, so the order dimension cannot increase during Stages 1 or 2.) It was proved by Hiraguchi in 1951 that $d\leq |P|/2$ holds for any finite poset. Thus, the order dimension of $(X,\preceq)$ is at most $2\cdot (|P|/2)=|P|$. On the other hand, the order dimension of the Boolean envelope of $(P,\succeq)$ is exactly $|P|$. This is another sense in which the poset $(X,\succeq)$ is smaller than or equal to the Boolean envelope

Let me describe a construction that meets your conditions. For this, let $(P,\succeq)$ be a poset (and always assume that $a\preceq b \Leftrightarrow b\succeq a$).

Stage 1. Let $(C,\succeq)$ be the Dedekind-MacNeille completion of $(P,\succeq)$, and let $h\colon P\to C\colon p\mapsto (p]$ be the order-embedding that takes an element $p\in P$ to its lower cone $(p] = \{x\in P\;|\;x\preceq p\}$.

Stage 2. Let $(D,\succeq)$ be the meet-subsemilattice $(C,\succeq)$ that is generated by the image $h(P)$. If this meet-semilattice has a least element, then label it $0$. If it does not have a least element, adjoin one and label it $0$.

At this point we have embedded $(P,\succeq)$ into a small meet-semilattice $(D,\succeq)$ with a zero element such that the maximal elements of $(P,\succeq)$ are mapped to the maximal elements of $(D,\succeq)$.

Stage 3. To each element of $d\in D$ that is not $0$ or an atom of $(D,\succeq)$ we shall adjoin a companion element $x_d$. Each new element $x_d$ will lie strictly below the element $d$ and also below any element $d'\in D$ for which $d'\succeq d$, $x_d$ will lie strictly above $0$, and $x_d$ will be incomparable with all other elements of $(D,\succeq)$. Let $(X,\succeq)$ be the poset obtained from $(D,\succeq)$ by adding all companion elements.

Stage 1 embeds $(P,\succeq)$ into a small complete lattice. Stage 2 ensures that Property 2 of the problem statement is satisfied. Stage 2 also ensures that maximal elements are preserved under the embedding. Stage 3 ensures that Property 1 of the problem statement is satisfied.

If you apply this construction to a chain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $2n-2$, which is much smaller than the size of the Boolean envelope of $P$. If you apply this construction to an antichain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $n+1$, which is also much smaller than the size of the Boolean envelope of $P$.

If $(P,\succeq)$ is a finite poset of order dimension $d$, then it can be shown that the order dimension of the poset $(X,\succeq)$ constructed above is at most $2d$. (To prove this for yourself, use the fact that the Dedekind-MacNeille completion does not increase order dimension, so the order dimension cannot increase during Stages 1 or 2.) It was proved by Hiraguchi in 1951 that $d\leq |P|/2$ holds for any finite poset. Thus, the order dimension of $(X,\succeq)$ is at most $2\cdot (|P|/2)=|P|$. On the other hand, the order dimension of the Boolean envelope of $(P,\succeq)$ is exactly $|P|$. This is another sense in which the poset $(X,\succeq)$ is smaller than or equal to the Boolean envelope

Let me describe a construction that meets your conditions. For this, let $(P,\succeq)$ be a poset (and always assume that $a\preceq b \Leftrightarrow b\succeq a$).

Stage 1. Let $(C,\succeq)$ be the Dedekind-MacNeille completion of $(P,\succeq)$, and let $h\colon P\to C\colon p\mapsto (p]$ be the order-embedding that takes an element $p\in P$ to its lower cone $(p] = \{x\in P\;|\;x\preceq p\}$.

Stage 2. Let $(D,\succeq)$ be the meet-subsemilattice $(C,\succeq)$ that is generated by the image $h(P)$. If this meet-semilattice has a least element, then label it $0$. If it does not have a least element, adjoin one and label it $0$.

At this point we have embedded $(P,\succeq)$ into a small meet-semilattice $(D,\succeq)$ with a zero element such that the maximal elements of $(P,\succeq)$ are mapped to the maximal elements of $(D,\succeq)$.

Stage 3. To each element of $d\in D$ that is not $0$ or an atom of $(D,\succeq)$ we shall adjoin a companion element $x_d$. Each new element $x_d$ will lie strictly below the element $d$ and also below any element $d'\in D$ for which $d'\succeq d$, $x_d$ will lie strictly above $0$, and $x_d$ will be incomparable with all other elements of $(D,\succeq)$. Let $(X,\succeq)$ be the poset obtained from $(D,\succeq)$ by adding all companion elements.

Stage 1 embeds $(P,\succeq)$ into a small complete lattice. Stage 2 ensures that Property 2 of the problem statement is satisfied. Stage 2 also ensures that maximal elements are preserved under the embedding. Stage 3 ensures that Property 1 of the problem statement is satisfied.

If you apply this construction to a chain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $2n-2$, which is much smaller than the size of the Boolean envelope of $P$. If you apply this construction to an antichain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $n+1$, which is also much smaller than the size of the Boolean envelope of $P$.

If $(P,\succeq)$ is a finite poset of order dimension $d$, then it can be shown that the order dimension of the poset $(X,\succeq)$ constructed above is at most $2d \;(\leq |P|)$. (To prove this for yourself, use the fact that the Dedekind-MacNeille completion does not increase order dimension, so the order dimension cannot increase during Stages 1 or 2.) It was proved by Hiraguchi in 1951 that $d\leq |P|/2$ holds for any finite poset. Thus, the order dimension of $(X,\preceq)$ is at most $2\cdot (|P|/2)=|P|$. On the other hand, the order dimension of the Boolean envelope of $(P,\succeq)$ is exactly $|P|$. This is another sense in which the poset $(X,\succeq)$ is smaller than or equal to the Boolean envelope

Let me describe a construction that meets your conditions. For this, let $(P,\succeq)$ be a poset (and always assume that $a\preceq b \Leftrightarrow b\succeq a$).

Stage 1. Let $(C,\succeq)$ be the Dedekind-MacNeille completion of $(P,\succeq)$, and let $h\colon P\to C\colon p\mapsto (p]$ be the order-embedding that takes an element $p\in P$ to its lower cone $(p] = \{x\in P\;|\;x\preceq p\}$.

Stage 2. Let $(D,\succeq)$ be the meet-subsemilattice $(C,\succeq)$ that is generated by the image $h(P)$. If this meet-semilattice has a least element, then label it $0$. If it does not have a least element, adjoin one and label it $0$.

At this point we have embedded $(P,\succeq)$ into a small meet-semilattice $(D,\succeq)$ with a zero element such that the maximal elements of $(P,\succeq)$ are mapped to the maximal elements of $(D,\succeq)$.

Stage 3. To each element of $d\in D$ that is not $0$ or an atom of $(D,\succeq)$ we shall adjoin a companion element $x_d$. Each new element $x_d$ will lie strictly below the element $d$ and also below any element $d'\in D$ for which $d'\succeq d$, $x_d$ will lie strictly above $0$, and $x_d$ will be incomparable with all other elements of $(D,\succeq)$. Let $(X,\succeq)$ be the poset obtained from $(D,\succeq)$ by adding all companion elements.

Stage 1 embeds $(P,\succeq)$ into a small complete lattice. Stage 2 ensures that Property 2 of the problem statement is satisfied. Stage 2 also ensures that maximal elements are preserved under the embedding. Stage 3 ensures that Property 1 of the problem statement is satisfied.

If you apply this construction to a chain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $2n-2$, which is much smaller than the size of the Boolean envelope of $P$. If you apply this construction to an antichain $(P,\succeq)$ of some length $n>1$, the resulting $(X,\succeq)$ has size $n+1$, which is also much smaller than the size of the Boolean envelope of $P$.

If $(P,\succeq)$ is a finite poset of order dimension $d$, then it can be shown that the order dimension of the poset $(X,\succeq)$ constructed above is at most $2d$. (To prove this for yourself, use the fact that the Dedekind-MacNeille completion does not increase order dimension, so the order dimension cannot increase during Stages 1 or 2.) It was proved by Hiraguchi in 1951 that $d\leq |P|/2$ holds for any finite poset. Thus, the order dimension of $(X,\preceq)$ is at most $2\cdot (|P|/2)=|P|$. On the other hand, the order dimension of the Boolean envelope of $(P,\succeq)$ is exactly $|P|$. This is another sense in which the poset $(X,\succeq)$ is smaller than or equal to the Boolean envelope