I will attempt to explain the distinction using the simplest example: the sample mean. Suppose we have an iid sample of random variables $\{X_i\}_{i=1}^n$. Then define the sample mean as $\bar{X}_n$. As the sample size grows, our value of the sample mean changes, hence the subscript $n$ to emphasize that our sample mean depends on the sample size.
Noting that $\bar{X}_n$ itself is a random variable, we can define a sequence of random variables, where elements of the sequence are indexed by different samples (sample size is growing), i.e. $\{\bar{X}_n\}_{n=1}^{\infty}$. The weak law of large numbers (WLLN) tells us that so long as $E(X_1^2)<\infty$, that
$$plim\bar{X}_n = \mu,$$
or equivalently
$$\bar{X}_n \rightarrow_P \mu,$$
where $\mu=E(X_1)$. Formally, convergence in probability is defined as
$$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. $$
In other words, the probability of our estimate being within $\epsilon$ from the true value tends to 1 as $n \rightarrow \infty$. Convergence in probability gives us confidence our estimators perform well with large samples.
Convergence in distribution tell us something very different and is primarily used for hypothesis testing. Under the same distributional assumptions described above, CLT gives us that
$$\sqrt{n}(\bar{X}_n-\mu) \rightarrow_D N(0,E(X_1^2)).$$
Convergence in distribution means that the cdf of the left-hand size converges at all continuity points to the cdf of the right-hand side, i.e.
$$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$
where $F_n(x)$ is the cdf of $\sqrt{n}(\bar{X}_n-\mu)$ and $F(x)$ is the cdf for a $N(0,E(X_1^2))$ distribution. Knowing the limiting distribution allows us to test hypotheses about the sample mean (or whatever estimate we are generating).
