Strict comma objects implies comma objects I'm condusion on a statement in this page comma object in $n$lab. It states:

any strict comma object is a comma object, but the converse is not in general true.

My confusion is: the strict comma object trivially satisfies the $1$-dimensional univerality, but how to show it satisfies the $2$-dimensional one?
 A: Your understanding of the definition is incorrect, but that is probably because the cited nLab page is misleading. Let me spell it out a little bit more accurately:

Let $\mathfrak{K}$ be a 2-category and let $f : A \to C$ and $g : B \to C$ be morphisms in $\mathfrak{K}$. A strict comma object $(f \downarrow g)$ in $\mathfrak{K}$ is an object $P$ in $\mathfrak{K}$ together with isomorphisms
  $$\mathfrak{K} (T, P) \cong (\mathfrak{K} (T, f) \downarrow \mathfrak{K} (T, g))$$
  that are 2-natural in $T$.

In particular, strict comma objects have a 1-dimensional universal property as well as a 2-dimensional universal property by definition. The 1-dimensional universal property is as stated on nLab: so we have a (not necessarily commutative!) diagram in $\mathfrak{K}$ of the form below,
$$\require{AMScd}
\begin{CD}
P @>{q}>> B \\
@V{p}VV @VV{g}V \\
A @>>{f}> C
\end{CD}$$
and a 2-cell $\alpha : f \circ p \Rightarrow g \circ q$ such that etc. The 2-dimensional universal property says exactly this:

Given morphisms $x_0, x_1 : T \to P$ and 2-cells $\phi : f \circ p \circ x_0 \Rightarrow f \circ p \circ x_1$ and $\psi : g \circ q \circ x_0 \Rightarrow g \circ q \circ x_1$ such that $\psi \bullet \alpha x_0 = \alpha x_1 \bullet \phi$, there is a unique 2-cell $\theta : x_0 \Rightarrow x_1$ such that $\phi = f p \theta$ and $\psi = g q \theta$.

This, you will notice, is what appears on nLab. So the difference between strict comma objects and (bicategorical) comma objects is elsewhere – in fact, it is in the 1-dimensional universal property: one has to replace the uniqueness clause with an appropriate up-to-isomorphism version so that the resulting notion is invariant under equivalences in $\mathfrak{K}$.
